Question

A line segment joining P(2, –3) and Q(0, –1) is cut by the x-axis at the point R. A line AB cuts the y-axis at T(0, 6) and is perpendicular to PQ at S.

Find the:

1. equation of line PQ
2. equation of line AB
3. coordinates of points R and S.

Answer 1

a. Slope of PQ = –1

Equation of PQ: x + y + 1 = 0

b. Slope of AB = 1

∴ Equation of line AB, x – y + 6 = 0

c. R(–1, 0)

`S(-7/2, 5/2)`

Answer 2

1. Equation of line PQ

`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`

(2, −3) = (x₁​, y_1​)

Q(0, −1) = (x₂​, y₂​)

Slope of PQ = m = `(-1-(-3))/(0-2) = 2/-2= -1`

y + 3 = −1(x − 2) ⇒ y + 3 = −x + 2 ⇒ y = −x −1

y = −x − 1

2. Coordinates of point R (where PQ cuts the x-axis)

0 = −x − 1 ⇒ x =−1 ⇒ R = (−1, 0)

(−1, 0)

3. Equation of line AB

Line AB cuts the y-axis at T(0, 6)

It is perpendicular to PQ

We already found:

Slope of PQ = −1

So slope of AB = reciprocal with opposite sign = 1

Now use point-slope form with slope 1 and point T(0, 6):

y − 6 = 1(x − 0) ⇒ y = x + 6

y = x + 6

4. Coordinates of point S (intersection of AB and PQ)

• y = −x − 1 (PQ)
• y = x + 6 (AB)

−x − 1 = x + 6 ⇒ −2x = 7 ⇒ x = `-7/2`

Now find y using any equation, say y = x + 6:

`y = -7/2+6=5/2`

`(-7/2, 5/2)`