**Question 6**:

The point

*M*is (–3, 5) and the point*N*is (4, 7). The point*P*moves such that*PM*:*PN*= 2: 3. Find the equation of the locus of*P*.

*Solution:*$\begin{array}{l}\text{Let}P=\left(x,y\right)\\ PM:PN=2:3\\ \frac{PM}{PN}=\frac{2}{3}\\ 3PM=2PN\\ 3\sqrt{{\left(x-\left(-3\right)\right)}^{2}+{\left(y-5\right)}^{2}}=2\sqrt{{\left(x-4\right)}^{2}+{\left(y-7\right)}^{2}}\end{array}$

Square both sides to eliminate the square roots.

9[

*x*^{2}+ 6*x*+ 9 +*y*^{2}– 10*y*+ 25] = 4 [*x*^{2}– 8*x*+ 16 +*y*^{2}– 14*y*+ 49]9

*x*^{2}+ 54*x +*9*y*^{2}– 90*y*+ 306 = 4*x*^{2}– 32*x*+ 4*y*^{2}– 56*y*+ 2605

*x*^{2}+ 5*y*^{2}+ 86*x*– 34*y*+ 46 = 0Hence, the equation of the locus of point

*P*is**5**

*x*^{2}+ 5*y*^{2}+ 86*x*– 34*y*+ 46 = 0

**Question 7**:

Given the points

*A*(0, 2) and*B*(6, 5). Find the equation of the locus of a moving point*P*such that the triangle*APB*always has a right angle at*P*.

*Solution:*Let

*P*= (*x, y*)Given that triangle

*APB*= 90^{o}, thus*AP*is perpendicular to*PB*.Hence, (

*m*)(_{AP}*m*) = –1._{PB}(

$\left(\frac{y-2}{x-0}\right)\left(\frac{y-5}{x-6}\right)=-1$

*m*)(_{AP}*m*) = –1_{PB}$\left(\frac{y-2}{x-0}\right)\left(\frac{y-5}{x-6}\right)=-1$

(

*y*– 2)(*y*– 5) = –*x*(*x*– 6)*y*

^{2}– 7

*y*+ 10 = –

*x*

^{2}+ 6

*x*

*y*

^{2}+

*x*

^{2}– 6

*x*– 7

*y*+ 10 = 0

Hence, the equation of the locus of point

*P*is

*y*^{2}+*x*^{2}– 6*x*– 7*y*+ 10 = 0.