Question

The coordinates of points on the line joining the points $P(3,-4)$ and $Q(-2,5)$ that is twice as from $P$ as from $Q$ is

• A. $(14,-7)$
• B. $(\frac{6}{7},\frac{19}{7})$
• C. $(-7,14)$
• D. $(7,14)$

Answer 1

The correct option is C $(-7,14)$

Let $A(x,y)$ be the required point.

Also, let a be distance of A from Q.

$\therefore$ Distance of A from P =2a

$\Rightarrow PQ=PA+AQ$

$=3a$

$\Rightarrow \sqrt{25+81}=3a$

$\sqrt{106}=3a$

$\Rightarrow \frac{A}{Q}.\sqrt{{(x-3)}^2+{(y+4)}^2}=2.\sqrt{{(x+2)}^2+{(y-5)}^2}$

$\Rightarrow {(x-3)}^2+{(y+4)}^2=4({(x+2)}^2+{(y-5)}^2)$

$x^2-6x+9+y^2-8y+16=4x^2+16x+16+4y^2-40y+100$

$3x^2+22x+7+3y^2-48y+84=0$

$3x^2+3y^2+22x-48y+91=0$

A lies on PQ

$\Rightarrow (x-3).\frac{9}{5}=(y+4)$

$\Rightarrow 9x-27=-5y-20$

$9x+5y-7=0$

$\Rightarrow x=\frac{7-5y}{9}$

$\therefore 3{\left(\frac{7-5y}{9}\right)}^2+3y^2+22\left(\frac{7-5y}{9}\right)-48y+91=0$

$\Rightarrow \frac{3}{81}(25y^2-70y+49)+3y^2+\frac{154}{9}-\frac{110}{9}y-48y+91=0$

$75y^2-210y+147+243y^2+1386-990y-3888y+7371=0$

$318y^2-5088y+8904=0$

$y=14,2$

$\Rightarrow x=-7,\frac{-1}{3}$

$\therefore A=(-7,14)$ or $(\frac{-1}{3},2)$

Hence, the answer is $(-7,14).$