Answer:
Q2. (16,8)
Q3. [tex]k=\frac{2}{3}[/tex], 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
[tex]|PQ|=|PR|[/tex]
This gives us:
[tex]\sqrt{(2a-2)^2+(a+5)^2}=\sqrt{(2a+3)^2+(6-a)^2}[/tex]
[tex]\implies (2a-2)^2+(a+5)^2=(2a+3)^2+(6-a)^2[/tex]
Expand:
[tex]4a^2-8a+4+a^2+10a+25=4a^2+12a+9+a^2 -12a+36[/tex]
[tex]2a=16[/tex]
[tex]a=8[/tex]
The coordinates of P are [tex](16,8)[/tex]
Q.3 The equation of the line segment joining the points
A (5.-6) and B (-1,-4) is [tex]x+3y=-13[/tex].
The x-coordinate of the point that divides AB in the ratio m:n is
[tex]x=\frac{mx_2+nx_1}{m+n}[/tex]
The y-axis meets this line at [tex](0,-\frac{13}{3})[/tex]
We substitute [tex]x_2=-1,x_1=5,x=0[/tex] into this equation and solve for m and n.
[tex]0=\frac{-m+5n}{m+n}[/tex]
[tex]m=5n[/tex]
[tex]\frac{m}{n}=\frac{5}{1}[/tex]
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 [tex]-7x+3y=23[/tex].
The point (-3, k) must satisfy this line because it lies on it.
[tex]-7(-3)+3k=23[/tex].
[tex]\implies k=\frac{2}{3}[/tex]
We again use the equation [tex]x=\frac{mx_2+nx_1}{m+n}[/tex] to find the given ratio.
Substitute: [tex]x_2=-2,x_1=-5[/tex]
[tex]4=\frac{-2m+-5n}{m+n}[/tex]
[tex]\implies m=2n[/tex]
[tex]\frac{m}{n}= \frac{2}{1}[/tex]
The ratio is m:n=2:1
Q. 5 The equation of the line joining A (2,3) and B(6,-3) is [tex]3x+2y=12[/tex].
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
