**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 (150

^{°}) ⇒ m = -1/√3, 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/√3 + 2

⇒ x – 2√3 + √3y = 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

⇒ 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-y

_{1}= m (x – x_{1}), hence the equation will be⇒ y – 0 = -2(x – (-3))

⇒ y = -2x -6

⇒ 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 ± 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.

^{-1}(3) with the x-axis and cuts off an intercept of 4 units on negative direction of y-axis.

**Solution:**

Here, ∅ = tan

^{-1}(3) ⇒ m = tan∅ =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 = (y

_{2}– y_{1}) / (x_{2}– x_{1}) = (2 – (-5) ) / (1 – 2 ) = -7Also, 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

⇒ 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

⇒ 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 30° with the positive direction of the x-axis.**

**Solution:**

Given, m = tan 30° = 1/√3

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 – x

_{1}), hence the equation will be⇒ y – 2 = 1/√3 . (x – 0)

⇒ √3y – 2√3 = x

⇒ x – √3y + 2√3 = 0

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.