Condition for a Point to Lie on a Line

For the line $$2x-y+5=0,$$

we have $$a=2,\qquad b=-1,\qquad c=5$$

If $$P(x_1,y_1)=(3,-2),$$

then its image in the line is

$$\left( x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2}, \ y_1-\frac{2b(ax_1+by_1+c) {a^2+b^2} \right)$$

Now,

$$ax_1+by_1+c = 2(3)-(-2)+5 =13$$

and

$$a^2+b^2=4+1=5$$

Hence image point is

$$\left( 3-\frac{2(2)(13)}5, \ -2-\frac{2(-1)(13)}5 \right)$$

$$= \left( 3-\frac{52}{5}, \ -2+\frac{26}{5} \right)$$

$$= \left( -\frac{37}{5}, \ \frac{16}{5} \right)$$

Since this lies on

$$x+\lambda y-7=0,$$

$$-\frac{37}{5}+\lambda\left(\frac{16}{5}\right)-7=0$$

$$-37+16\lambda-35=0$$

$$16\lambda=72$$

$$\lambda=\frac92$$