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Chapter 14

Equation of a Straight Line — Assertion-Reason Type Questions

Class - 10 RS Aggarwal Mathematics Solutions



Assertion-Reason Type Questions

Question 1

Assertion (A): The slope of the line passing through the points (3, -2) and (-7, -2) is 0.

Reason (R): The gradient (slope) of a line passing through the points (x1, y1) and (x2, y2) is x2x1y2y1\dfrac{x_2 - x_1}{y_2 - y_1}.

options

  1. Both A and R are true, and R is the correct explanation of A.

  2. Both A and R are true, but R is not the correct explanation of A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Slope = y2y1x2x1=2(2)73=010=0\dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-2 - (-2)}{-7 - 3} = \dfrac{0}{-10} = 0.

So, Assertion (A) is true.

The correct slope formula is y2y1x2x1\dfrac{y_2 - y_1}{x_2 - x_1}, whereas the Reason gives x2x1y2y1\dfrac{x_2 - x_1}{y_2 - y_1}, which is incorrect.

So, Reason (R) is false.

Hence, Option 3 is the correct option.

Question 2

Assertion (A): The angle of inclination of the line y = x3\dfrac{x}{\sqrt{3}} - 5 is 60°.

Reason (R): The gradient m of a line is given by m = tan θ.

options

  1. Both A and R are true, and R is the correct explanation of A.

  2. Both A and R are true, but R is not the correct explanation of A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Comparing y=13x5y = \dfrac{1}{\sqrt{3}}x - 5 with y = mx + c, we get m=13m = \dfrac{1}{\sqrt{3}}.

tanθ=13tanθ=tan30°θ=30°.\Rightarrow \tan \theta = \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow \tan \theta = \tan 30° \\[1em] \Rightarrow \theta = 30°.

The inclination is 30°, not 60°, so Assertion (A) is false.

The gradient m of a line is given by m = tan θ, where θ is the angle of inclination, so Reason (R) is true.

Hence, Option 4 is the correct option.

Question 3

Assertion (A): The slope of the line perpendicular to the line passing through the points (2, 5) and (-3, 6) is given by 5.

Reason (R): The product of the slopes of two perpendicular lines is always -1.

options

  1. Both A and R are true, and R is the correct explanation of A.

  2. Both A and R are true, but R is not the correct explanation of A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Slope of the line through (2, 5) and (-3, 6) :

m1=6532=15.\Rightarrow m_1 = \dfrac{6 - 5}{-3 - 2} = -\dfrac{1}{5}.

The product of the slopes of two perpendicular lines is -1. Let the slope of the perpendicular line be m2.

m1×m2=115×m2=1m2=5.\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{1}{5} \times m_2 = -1 \\[1em] \Rightarrow m_2 = 5.

So, Assertion (A) is true and Reason (R) is true, and the Reason is the principle used to obtain the slope in the Assertion.

Hence, Option 1 is the correct option.

Question 4

Assertion (A): The equation of the line whose inclination is 45° and which intersects the y-axis at the point (0, -4) is x - y = 4.

Reason (R): The equation of the line having slope m and y-intercept c is given by y = cx + m.

options

  1. Both A and R are true, and R is the correct explanation of A.

  2. Both A and R are true, but R is not the correct explanation of A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Slope m = tan 45° = 1, and the y-intercept is c = -4.

By slope-intercept form,

y=mx+cy=(1)x+(4)y=x4xy=4.\Rightarrow y = mx + c \\[1em] \Rightarrow y = (1)x + (-4) \\[1em] \Rightarrow y = x - 4 \\[1em] \Rightarrow x - y = 4.

So, Assertion (A) is true.

The correct slope-intercept form is y = mx + c, whereas the Reason gives y = cx + m, which is incorrect.

So, Reason (R) is false.

Hence, Option 3 is the correct option.

Question 5

Assertion (A): A line is parallel to the line 2x - 3y = 7 and it passes through the point (0, 4). The equation of this line is 2x - 7y + 28 = 0.

Reason (R): Slope of the line y = mx + c is m.

options

  1. Both A and R are true, and R is the correct explanation of A.

  2. Both A and R are true, but R is not the correct explanation of A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

The line 2x - 3y = 7 can be written as y=23x73y = \dfrac{2}{3}x - \dfrac{7}{3}, so its slope is 23\dfrac{2}{3}.

The required line is parallel to it, so its slope is 23\dfrac{2}{3}, and it passes through (0, 4), so c = 4.

By slope-intercept form,

y=23x+43y=2x+122x3y+12=0.\Rightarrow y = \dfrac{2}{3}x + 4 \\[1em] \Rightarrow 3y = 2x + 12 \\[1em] \Rightarrow 2x - 3y + 12 = 0.

The correct equation is 2x - 3y + 12 = 0, not 2x - 7y + 28 = 0, so Assertion (A) is false.

The slope of the line y = mx + c is indeed m, so Reason (R) is true.

Hence, Option 4 is the correct option.

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