Question

15 The straight lines 3x + 4y - 5 = 0 and 4x = 3y + 15
intersect at the point P. On these lines the points Q
and R are chosen so that PQ = PR. The slopes of the
lines QR passing through (1, 2) are
a. -7,1/7 b. 7,1/7 c. 7,-1/7 d. 3,-1/3

Answer 1

Answer:

1) -7, 1/7

2) 7, 1/7

3) 7, (-1/7)

4) 3, -(1/3)

Solution:

Given lines are 3x + 4y – 5 = 0…(i)

4x = 3y + 15…(ii)

Slope of (i) m1 = -¾

Slope of (ii) m2 = 4/3

Here m1m2 = -1

So the lines are perpendicular.

Given PQ = PR

So triangle PQR is isosceles.

Let slope of QR = m

Then tan θ = |(m2 – m)/(1+m2m)|

tan 450 = |(4/3) – m)/(1+(4/3)m)|

1 = ±(4 – 3m)/(3+ 4m)

3 + 4m = 4 – 3m

7m = 1

So m = 1/7

When 3 + 4m = -(4 – 3m)

3 + 4m + 4 – 3m = 0

m = -7

=> m = 1/7, -7

Hence option (1) is the answer.

Answer 2

Answer:

Given lines are 3x + 4y – 5 = 0…(i)

4x = 3y + 15…(ii)

Slope of (i) m1 = -¾

Slope of (ii) m2 = 4/3

Here m1m2 = -1

So the lines are perpendicular.

Given PQ = PR

So triangle PQR is isosceles.