RS Aggarwal Class 8 Mathematics Solutions Chapter 25 Graphs

Question 1. Identify the coordinates of the points shown on the coordinate plane and state which quadrant each lies in.
Answer: The coordinate plane shows several marked points. Point (-2, 2) lies in Quadrant II where the x-coordinate is negative and the y-coordinate is positive. Point (4, 5) is located in Quadrant I where both coordinates are positive. Point (-3, -4) is in Quadrant III where both x and y coordinates are negative. Point (5, -6) falls in Quadrant IV where the x-coordinate is positive but the y-coordinate is negative. The origin (0, 0) is the intersection of both axes.
In simple words: Each point on the plane has two numbers - one for left/right (x) and one for up/down (y). Depending on whether these numbers are positive or negative, the point lands in one of four quadrants around the origin.

Exam Tip: Always check the signs of both coordinates carefully - positive x goes right, negative x goes left; positive y goes up, negative y goes down. This determines which quadrant the point occupies.

Question 2. (i) On the x - axis, take 4 units to the right of the y axis; and then on the y - axis, take 3 units above the x - axis. Thus, identify the point obtained.
Answer: Moving 4 units to the right along the x-axis means the x-coordinate is 4. Moving 3 units upward along the y-axis means the y-coordinate is 3. Therefore, the point obtained is A(4, 3).
In simple words: Start at the origin, go 4 steps right, then go 3 steps up. The point where you land has coordinates (4, 3).

Exam Tip: Always move horizontally first (along x-axis), then vertically (along y-axis). The horizontal distance gives the x-coordinate, and the vertical distance gives the y-coordinate.

Question 2. (ii) On the x - axis, take 2 units to the right of the y axis; and then on the y - axis, take 6 units above the x - axis. Thus, identify the point obtained.
Answer: Moving 2 units to the right along the x-axis means the x-coordinate is 2. Moving 6 units upward along the y-axis means the y-coordinate is 6. Therefore, the point obtained is B(2, 6).
In simple words: Start at the origin, go 2 steps right, then go 6 steps up. The point where you land has coordinates (2, 6).

Exam Tip: Remember that "above the x - axis" means positive y values, and "to the right" means positive x values.

Question 2. (iii) On the x - axis, take 3 units to the left of the y axis; and then on the y - axis, take 5 units above the x - axis. Thus, identify the point obtained.
Answer: Moving 3 units to the left along the x-axis means the x-coordinate is -3. Moving 5 units upward along the y-axis means the y-coordinate is 5. Therefore, the point obtained is C(-3, 5).
In simple words: Start at the origin, go 3 steps left, then go 5 steps up. The point where you land has coordinates (-3, 5).

Exam Tip: When moving left on the x-axis, the x-coordinate becomes negative. Always be careful with the sign when the direction is "left" or "below".

Question 2. (iv) On the x - axis, take 5 units to the left of the y axis; and then on the y - axis, take 2 units above the x - axis. Thus, identify the point obtained.
Answer: Moving 5 units to the left along the x-axis means the x-coordinate is -5. Moving 2 units upward along the y-axis means the y-coordinate is 2. Therefore, the point obtained is D(-5, 2).
In simple words: Start at the origin, go 5 steps left, then go 2 steps up. The point where you land has coordinates (-5, 2).

Exam Tip: The larger the leftward movement, the more negative the x-coordinate becomes. Pay attention to the distance value given in the problem.

Question 2. (v) On the x - axis, take 2 units to the left of the y axis; and then on the y - axis, take 3 units below the x - axis. Thus, identify the point obtained.
Answer: Moving 2 units to the left along the x-axis means the x-coordinate is -2. Moving 3 units downward along the y-axis means the y-coordinate is -3. Therefore, the point obtained is E(-2, -3).
In simple words: Start at the origin, go 2 steps left, then go 3 steps down. The point where you land has coordinates (-2, -3).

Exam Tip: "Below the x - axis" means a negative y-coordinate. Both negative values place this point in Quadrant III.

Question 2. (vi) On the x - axis, take 5 units to the left of the y axis; and then on the y - axis, take 3 units below the x - axis. Thus, identify the point obtained.
Answer: Moving 5 units to the left along the x-axis means the x-coordinate is -5. Moving 3 units downward along the y-axis means the y-coordinate is -3. Therefore, the point obtained is F(-5, -3).
In simple words: Start at the origin, go 5 steps left, then go 3 steps down. The point where you land has coordinates (-5, -3).

Exam Tip: When both coordinates are negative, the point is always in Quadrant III. Double-check the directions given - left and below both produce negative values.

Question 2. (vii) On the x - axis, take 5 units to the right of the y axis; and then on the y - axis, take 4 units below the x - axis. Thus, identify the point obtained.
Answer: Moving 5 units to the right along the x-axis means the x-coordinate is 5. Moving 4 units downward along the y-axis means the y-coordinate is -4. Therefore, the point obtained is G(5, -4).
In simple words: Start at the origin, go 5 steps right, then go 4 steps down. The point where you land has coordinates (5, -4).

Exam Tip: A positive x and negative y places the point in Quadrant IV. Remember that "right" is positive and "down" is negative.

Question 2. (viii) On the x - axis, take 3 units to the right of the y axis; and then on the y - axis, take 3 units below the x - axis. Thus, identify the point obtained.
Answer: Moving 3 units to the right along the x-axis means the x-coordinate is 3. Moving 3 units downward along the y-axis means the y-coordinate is -3. Therefore, the point obtained is H(3, -3).
In simple words: Start at the origin, go 3 steps right, then go 3 steps down. The point where you land has coordinates (3, -3).

Exam Tip: When the horizontal and vertical distances are equal, the point lies on a line at a 45-degree angle. This point is in Quadrant IV.

Question 3. (a) The given function is y = 3x. For some different values of x, the corresponding values of y are given. Plot the points O(0, 0), A(1, 3) and B(2, 6). Join them successively to obtain the required graph.
Answer: For the linear equation y = 3x, substitute the given x values to find corresponding y values. When x = 0, y = 3(0) = 0, giving point O(0, 0). When x = 1, y = 3(1) = 3, giving point A(1, 3). When x = 2, y = 3(2) = 6, giving point B(2, 6). Plot these three points on the coordinate plane. Connect them with a straight line, which represents the graph of y = 3x. This line passes through the origin and has a slope of 3, meaning for every 1 unit moved right, the line goes up 3 units.
In simple words: The equation y = 3x means y is always three times larger than x. When you plot the points and connect them, you get a straight line that tilts upward from left to right.

Exam Tip: For linear equations, plotting just two points is technically enough to draw the line, but using three points helps verify accuracy. Always check that your points satisfy the given equation before plotting.

Question 3. (b) Reading off from the graph: (i) On the x - axis, take the point L at x = 3. Draw LP \( \perp \) x - axis, meeting the graph at P. Clearly, PL = 9 units. Therefore, x = 3 \( \implies \) y = 9.
Answer: To read values from the graph, locate the given x-value on the horizontal axis. At x = 3, draw a vertical line perpendicular to the x-axis. This perpendicular line meets the graph at point P. Measure the vertical distance from the x-axis to point P, which is 9 units. This means the y-coordinate is 9. Therefore, when x = 3, the corresponding value of y is 9, confirming the relationship y = 3x.
In simple words: Find the x-value on the bottom axis, draw a line straight up to hit the graph line, then read how high up it goes. That height is your y-value.

Exam Tip: Always draw the perpendicular lines carefully - horizontal lines to find x, vertical lines to find y. Use a ruler to ensure accuracy when reading coordinates from the graph.

Question 3. (b) (ii) On the x - axis, take the point M at x = 5. Draw MQ \( \perp \) x - axis, meeting the graph at Q. Clearly, QM = 15 units. Therefore, x = 5 \( \implies \) y = 15.
Answer: Locate x = 5 on the x-axis and label it as point M. Draw a perpendicular line from M upward to intersect the graph at point Q. Measure the vertical distance from the x-axis to Q, which equals 15 units. This vertical measurement represents the y-coordinate. Therefore, when x = 5, we get y = 15, which satisfies the equation y = 3x since 3 times 5 equals 15.
In simple words: At x = 5, the vertical distance up to the graph line is 15 units. So when x = 5, y = 15.

Exam Tip: Notice the pattern - as x increases, y increases proportionally. For this linear equation, every increase in x by 1 unit causes y to increase by 3 units.

Question 3. (b) (iii) On the x - axis, take the point N at x = 6. Draw RN \( \perp \) x - axis, meeting the graph at R. Clearly, RN = 18 units. Therefore, x = 6 \( \implies \) y = 18.
Answer: Identify the point N at x = 6 on the x-axis. Draw a perpendicular line from N upward until it touches the graph at point R. The vertical distance from the x-axis to R measures 18 units, which is the y-coordinate. Therefore, when x = 6, the corresponding value is y = 18, confirming the linear relationship y = 3x.
In simple words: At x = 6, go straight up to the line and measure the height. The height is 18 units, so y = 18.

Exam Tip: Reading values from a graph requires precision. Use grid lines as guides and estimate carefully when the point falls between grid lines.

Question 2. (a) The given function is P = 4x. For some different values of x, the corresponding values of P are given. Plot the points O(0, 0), A(1, 4) and B(2, 8). Join them successively to obtain the required graph.
Answer: For the linear equation P = 4x, substitute values of x to find the corresponding P values. When x = 0, P = 4(0) = 0, producing point O(0, 0). When x = 1, P = 4(1) = 4, producing point A(1, 4). When x = 2, P = 4(2) = 8, producing point B(2, 8). Plot all three points on the coordinate plane and connect them with a straight line, which represents the graph of P = 4x. This line passes through the origin with a slope of 4.
In simple words: The relationship P = 4x means P is always four times as large as x. When you plot and join the three points, you create a straight line that slopes upward steeply.

Exam Tip: Notice that the slope is steeper here (4) compared to the previous graph (slope 3). A larger slope value makes the line tilt more steeply upward.

Question 2. (b) Reading off from the graph: (i) On the x - axis, take the point L at x = 3. Draw LR \( \perp \) x - axis, meeting the graph at R. Clearly, RL = 12 units. Therefore, x = 3 \( \implies \) P = 12.
Answer: Locate the point L at x = 3 on the x-axis. Draw a vertical line from L perpendicular to the x-axis, meeting the graph at point R. The vertical distance from the x-axis to R equals 12 units, representing the P-coordinate. Therefore, when x = 3, we get P = 12, which satisfies P = 4x since 4 times 3 equals 12.
In simple words: At x = 3, the vertical distance up to the graph line is 12 units. So P = 12 when x = 3.

Exam Tip: Always verify your graph readings by substituting back into the original equation. If the values don't match, check your plotting or reading technique.

Question 2. (b) (ii) On the x - axis, take the point M at x = 4. Draw MS \( \perp \) x - axis, meeting the graph at S. Clearly, SM = 16 units. Therefore, x = 4 \( \implies \) P = 16.
Answer: Mark point M at x = 4 on the x-axis. Draw a perpendicular line from M upward to touch the graph at point S. Measure the vertical distance from the x-axis to S, which is 16 units. This represents the P-value. Therefore, when x = 4, the corresponding value is P = 16, confirming the relationship P = 4x.
In simple words: At x = 4, go straight up to the line and read the height. The height is 16 units, so P = 16.

Exam Tip: For each x-value on the graph, there is exactly one corresponding P-value. This unique pairing is what makes this a function.

Question 2. (b) (iii) On the x - axis, take the point N at x = 6. Draw NT \( \perp \) x - axis, meeting the graph at T. Clearly, TN = 24 units. Therefore, x = 6 \( \implies \) P = 24.
Answer: Find point N at x = 6 on the x-axis. Draw a perpendicular line from N upward until it reaches the graph at point T. The vertical measurement from the x-axis to T is 24 units, giving the P-coordinate. Therefore, when x = 6, we obtain P = 24, which matches the equation P = 4x since 4 times 6 equals 24.
In simple words: At x = 6, the vertical distance to the graph line is 24 units. So P = 24 when x = 6.

Exam Tip: Graph reading becomes more reliable when you use the grid lines carefully. Mark your x-position clearly and draw a perfectly vertical line to find the corresponding y or P value.

Question 3. (a) The given function is A = x². For some different values of x, the corresponding values of A are given. Plot the points O(0, 0), S(1, 1) and P(2, 4). Join them successively to obtain the required graph.
Answer: For the quadratic equation A = x², substitute the given x values to find corresponding A values. When x = 0, A = 0² = 0, giving point O(0, 0). When x = 1, A = 1² = 1, giving point S(1, 1). When x = 2, A = 2² = 4, giving point P(2, 4). Plot these three points on the coordinate plane. Connect them with a smooth curve (not straight lines), which represents the graph of A = x². This creates a parabola opening upward with its vertex at the origin.
In simple words: The equation A = x² means A equals x multiplied by itself. When you plot these points and join them with a curved line, you get a parabola that opens upward like a smile.

Exam Tip: For quadratic equations, connect the points with a smooth curve, not straight line segments. This curve is called a parabola. The shape becomes a key indicator that the relationship is quadratic, not linear.

Question 3. (b) Reading off from the graph: (i) On the x - axis, take the point L at x = 2. Draw LP \( \perp \) x - axis, meeting the graph at P. Clearly, PL = 4 units. Therefore, x = 2 \( \implies \) A = 4.
Answer: Locate the point L at x = 2 on the x-axis. Draw a perpendicular line from L upward to intersect the parabola at point P. Measure the vertical distance from the x-axis to P, which is 4 units. This measurement gives the A-coordinate. Therefore, when x = 2, the corresponding value is A = 4, which satisfies A = x² since 2² equals 4.
In simple words: At x = 2, the vertical distance up to the curve is 4 units. So A = 4 when x = 2.

Exam Tip: When reading from a curved graph, be extra careful to locate where your perpendicular line actually touches the curve, not just near it. A small error in placement causes significant reading errors.

Question 3. (b) (ii) On the x - axis, take the point M at x = 3. Draw MQ \( \perp \) x - axis, meeting the graph at Q. Clearly, QM = 9 units. Therefore, x = 3 \( \implies \) A = 9.
Answer: Mark point M at x = 3 on the x-axis. Draw a vertical line from M perpendicular to the x-axis until it meets the curve at point Q. The vertical distance from the x-axis to Q measures 9 units, which is the A-coordinate. Therefore, when x = 3, the corresponding value is A = 9, confirming the quadratic relationship A = x².
In simple words: At x = 3, go straight up to the curve and measure the height. The height is 9 units, so A = 9.

Exam Tip: Quadratic relationships increase more rapidly than linear ones as x increases. Notice how A jumps from 4 to 9 as x goes from 2 to 3 - this acceleration of growth is typical of quadratic functions.

Question 3. (b) (iii) On the x - axis, take the point N at x = 4. Draw RN \( \perp \) x - axis, meeting the graph at R. Clearly, RN = 16 units. Therefore, x = 4 \( \implies \) A = 16.
Answer: Identify point N at x = 4 on the x-axis. Draw a perpendicular line from N upward to touch the parabola at point R. Measure the vertical distance from the x-axis to R, which equals 16 units. This distance represents the A-coordinate. Therefore, when x = 4, we obtain A = 16, which satisfies A = x² since 4² equals 16.
In simple words: At x = 4, the vertical distance up to the curve is 16 units. So A = 16 when x = 4.

Exam Tip: As you move further from the origin along the x-axis, the parabola rises more steeply. This steepening of the curve is characteristic of quadratic relationships.

Question 1. Since the signs of coordinates are (+ , +), the point P(3, 6) lies in the I quadrant.
Answer: When both the x-coordinate and y-coordinate of a point are positive, the point is located in Quadrant I (the upper right region of the coordinate plane). Since P has coordinates (3, 6), where 3 is positive and 6 is positive, point P lies in Quadrant I.
In simple words: Quadrant I is the top-right section where both numbers are plus. If your point's numbers are both positive, it's in Quadrant I.

Exam Tip: Memorize the sign pattern for each quadrant - Quadrant I is (+ , +), II is (- , +), III is (- , -), and IV is (+ , -). This speeds up quadrant identification.

Question 2. Since the signs of coordinates are (- , -), the point (-7, -1) lies in the III quadrant.
Answer: When both the x-coordinate and y-coordinate of a point are negative, the point is positioned in Quadrant III (the lower left region of the coordinate plane). Since this point has coordinates (-7, -1), where both - 7 and - 1 are negative values, the point lies in Quadrant III.
In simple words: Quadrant III is the bottom-left section where both numbers are minus. If your point's numbers are both negative, it's in Quadrant III.

Exam Tip: Quadrant III is diagonally opposite to Quadrant I. Points where both coordinates are negative always end up in this lower-left region.

Question 3. Since the signs of the coordinates are (+ , -), the point A(2, -3) lies in the IV quadrant.
Answer: When the x-coordinate is positive and the y-coordinate is negative, the point is found in Quadrant IV (the lower right region of the coordinate plane). Since A has coordinates (2, -3), where 2 is positive and -3 is negative, point A lies in Quadrant IV.
In simple words: Quadrant IV is the bottom-right section where x is plus and y is minus. This region is below the x-axis but to the right of the y-axis.

Exam Tip: Think of Quadrant IV as the lower right - positive to the right, negative downward. Many students confuse this with Quadrant II, so always double-check the sign of each coordinate.

Question 4. Since the signs of coordinates are (- , +), the point Q(-4, 1) lies in the II quadrant.
Answer: When the x-coordinate is negative and the y-coordinate is positive, the point is located in Quadrant II (the upper left region of the coordinate plane). Since Q has coordinates (-4, 1), where -4 is negative and 1 is positive, point Q lies in Quadrant II.
In simple words: Quadrant II is the top-left section where x is minus and y is plus. This region is above the x-axis but to the left of the y-axis.

Exam Tip: Quadrant II is diagonally opposite to Quadrant IV. If one coordinate is negative and the other positive, you're in either Quadrant II or IV - check whether you're left (II) or right (IV).

Question 5. The y - axis is the abscissa of a point is its distance from the y - axis.
Answer: The abscissa of a point refers to its x-coordinate, which represents the horizontal distance measured from the y-axis. In other words, the abscissa tells you how far left or right the point is from the y-axis.
In simple words: The abscissa is just another name for the x-coordinate, or how far the point is from the y-axis in the left-right direction.

Exam Tip: Remember that abscissa = x-coordinate (distance from y-axis), and ordinate = y-coordinate (distance from x-axis). These are older mathematical terms sometimes appearing in geometry problems.

Question 6. A line parallel to the x - axis: The graph of y = a is a line parallel to the x - axis.
Answer: When an equation has the form y = a, where a is a constant (a fixed number), the resulting graph is a horizontal line that runs parallel to the x-axis. All points on this line have the same y-coordinate value of a, but their x-coordinates can be any value.
In simple words: If y always equals the same number, like y = 5, then the graph is a flat horizontal line at that height, running left and right forever.

Exam Tip: Equations of the form y = constant produce horizontal lines, while equations of the form x = constant produce vertical lines. These are special cases worth memorizing.

Question 7. The equation representing the y - axis is x = 0.
Answer: The y-axis itself can be described by the equation x = 0. Every point on the y-axis has an x-coordinate of zero, while the y-coordinate can be any value. Therefore, x = 0 is the defining equation for the y-axis.
In simple words: The y-axis is the vertical line where every point has x = 0. All points on this line are zero units away from it in the horizontal direction.

Exam Tip: Similarly, the x-axis has the equation y = 0. Both axes are special lines where one coordinate is always zero. Knowing these basic equations saves time in coordinate geometry problems.