Question

find the values of p for which the following quadratic equation has two equal roots:(p-12)x2 + 2(p-12)x + 2 = 0

Answer 1

(P -12)x² +2( P -12) x + 2 =0 have two real and equal roots so,
D = b² -4ac =0

here,
b = 2( P -12)
a = (P -12)
c = 2
now,
D = {2(P -12)}² -4.2(P -12) =0
4{ (P -12)² -2( P -12) } =0
(P -12)² -2(P -12)=0
(P -12){ P -12 -2} = 0
(P -12)( P -14) =0
P = 12, 14

hence value of P = 12, 14

Answer 2

A(2,2) and B(5,7) are the given points.

We know that the slope of the line through the points (x1,x2)and(y1,y2) is y2-y1 / x2-x1

Then the slope of the line AB is m1 = 7-2 / 5-2 = 5/3.

The given eqn of the other line is: 3x + Py - 9 = 0 ---(1)

We know that the slope of the line ax+by+c=0 is "-co-efficient of x / co-efficient of y".

then the slope of the line 3x + Py - 9 = 0 is m2 = -3/P.

For two lines to be perpendicular, the product of their slopes should be -1.

i.e., m1.m2 = -1

5/3 x -3/P = -1.

-5/P = -1

P = 5

Answer to this question is 5