Question

Q.2

Answer 1

Answer:

Q2. (16,8)

Q3. $k=\frac{2}{3}$, ratio=5:1

Q4. Ratio=2:1

Q5. Ratio=1:1

Step-by-step explanation:

Q2. Let (2a,a) be the coordinates of P.

Since P is equidistant  from Q (2,-5) and R (-3, 6), we have

$|PQ|=|PR|$

This gives us:

$\sqrt{(2a-2)^2+(a+5)^2}=\sqrt{(2a+3)^2+(6-a)^2}$

$\implies (2a-2)^2+(a+5)^2=(2a+3)^2+(6-a)^2$

Expand:

$4a^2-8a+4+a^2+10a+25=4a^2+12a+9+a^2 -12a+36$

$2a=16$

$a=8$

The coordinates of P are $(16,8)$

Q.3  The equation of the line segment joining the points

A (5.-6) and B (-1,-4) is $x+3y=-13$.

The x-coordinate of the point that divides AB in the ratio m:n is

$x=\frac{mx_2+nx_1}{m+n}$

The y-axis meets this line at $(0,-\frac{13}{3})$

We substitute $x_2=-1,x_1=5,x=0$ into this equation and solve for m and n.

$0=\frac{-m+5n}{m+n}$

$m=5n$

$\frac{m}{n}=\frac{5}{1}$

Therefore the ratio is m:n=5:1

Q.4 The equation of the line segment joining

the points (-5,-4) and (-2,3) is $-7x+3y=23$.

The point (-3, k) must satisfy this line because it lies on it.

$-7(-3)+3k=23$.

$\implies k=\frac{2}{3}$

We again use the equation $x=\frac{mx_2+nx_1}{m+n}$ to find the given ratio.

Substitute: $x_2=-2,x_1=-5$

$4=\frac{-2m+-5n}{m+n}$

$\implies m=2n$

$\frac{m}{n}= \frac{2}{1}$

The ratio is m:n=2:1

Q. 5 The equation of the line joining A (2,3) and B(6,-3) is $3x+2y=12$.

We substitute (4,m) to get:

12+4m=12

4m=0

m=0

It is obvious that: (4,0) is the midpoint of A(2,3) and B(6,-3).

Hence the ratio is 1:1