# Class 11 RD Sharma Solutions- Chapter 23 The Straight Lines- Exercise 23.3

### Question 1. Find the equation of a line making an angle of 150 degrees with the x-axis and cutting off an intercept 2 from y-axis.

Solution:

Given, slope, m = tan (150Â°) â‡’ m = -1/âˆš3, also, y-intercept = (0,2)

We know that equation, of a line is given as y = mx+ c, m is the slope and c is the intercept that line cuts on y-axis, therefore

equation for the line will be:

y =  -x/âˆš3 + 2

â‡’ x – 2âˆš3 + âˆš3y = 0

### Question 2. Find the equation of the straight line:

(i) with slope 2 and y-intercept 3;

(ii) with slope -1/3 and y-intercept -4

(iii) with slope -2 and intersecting the x-axis at a distance 3 units to the left of origin.

Solution:

(i) We know that equation, of a line is given as y = mx+ c, therefore equation for the line with slope 2 and y-intercept 3 will be: y = 2x + 3

(ii) Similarly, equation of the line with slope -1/3 and y-intercept -4 will be: y = -x/3 – 4

â‡’ x +3y +12 = 0

(iii) Since, the line cuts the x-axis at a distance 3 units to the left of origin its coordinate will be (-3,0) and the given slope, m = -2.

Equation of a line passing through a point is given by the formula: y-y1  = m (x – x1), hence the equation will be

â‡’ y – 0 = -2(x – (-3))

â‡’ y = -2x -6

â‡’ 2x + y + 6 = 0

### Question 3. Find the equations of the bisectors of the angle between the coordinate axes.

Solution:

The equation of the line on the coordinate axes are x=0 and y=0.

The equations of the bisectors of the angle between  x=0 and y=0 are:

x Â± y = 0

### Question 4. Find the equation of a line which makes an angle of tan-1(3) with the x-axis and cuts off an intercept of 4 units on negative direction of y-axis.

Solution:

Here, âˆ… = tan-1 (3) â‡’ m = tanâˆ… =3, line cuts cuts off an intercept of 4 units on negative direction of y-axis, so the coordinate will be (0, -4)

Hence, the equation of the line is: y = 3x -4

### Question 5. Find the equation of a line that has y-intercept -4 and is parallel to the line joining (2, -5) and (1,2).

Solution:

Since, our required line is parallel to the line passing through the coordinates (2, -5) and (1,2), it will have the same slope as the later line. Therefore, slope, m = (y2 –  y1) / (x2 – x1) = (2 – (-5) ) / (1 – 2 ) = -7

Also, given y-intercept = -4, hence the required equation of the line is: y = -7x – 4

### Question 6. Find the equation of a line which is perpendicular to the line joining (4,2) and (3,5) and cuts off an intercept of length 3 on y-axis.

Solution:

Slope of the line passing through the points (4,2) and (3,5) is

(y2 –  y1) / (x2 – x1) = (5 – 2 ) / (3 – 4 ) = -3

Now, our required line is perpendicular to the former line, so its slope will be m = 1/3

Also, y-intercept, c = 3, hence the required equation of the line is: y = x/3 + 3

â‡’ x – 3y +9 = 0

### Question 7. Find the equation of the perpendicular to the line (4,3) and (-1,1) if it cuts off an intercept -3 from y-axis.

Solution:

Slope of the line passing through the points (4,3) and (-1,1) is

(y2 –  y1) / (x2 – x1) = (1 – 3 ) / (-1 – 4 ) = 2/5

Now, our required line is perpendicular to the former line, so its slope will be m = -5/2

Also, y-intercept, c = -3, hence the required equation of the line is: y = -5x/2 – 3

â‡’ 5x + 2y +6 = 0

### Question 8. Find the equation of the straight line intersecting y-axis at a distance of 2 units above the origin and making an angle 30Â° with the positive direction of the x-axis.

Solution:

Given, m = tan 30Â° = 1/âˆš3

Since, the line intersects the y-axis at a distance 2 units above the origin its coordinate will be (0,2)

Equation of a line passing through a point is given by the formula: y-y1  = m (x – x1), hence the equation will be

â‡’ y – 2 = 1/âˆš3 . (x – 0)

â‡’ âˆš3y – 2âˆš3 = x

â‡’ x – âˆš3y + 2âˆš3 = 0

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