Find the coordinates of the point on the y-axis which is equidistant from (5,2) and (−4,3).
(0,2)
(0,-2)
(0,5)
(0,-5)
Explanation: Let $P(0,y)$. Then $PA^{2}=(0-5)^{2}+(y-2)^{2}=y^{2}-4y+29$ and $PB^{2}=(0+4)^{2}+(y-3)^{2}=y^{2}-6y+25$. Setting $PA^{2}=PB^{2}$: $-4y+29=-6y+25 \Rightarrow 2y=-4 \Rightarrow y=-2$. The point is $(0,-2)$.
Question 2 of 6medium
A(2,3) and B(5,7) are two points. Find the point P on AB such that AP:PB = 3:2.
The vertices of a triangle are (t,2t), (2t,3t), (3t,5t). Find its area in terms of t.
0
\(t^{2}\)/2
\(t^{2}\)
\(2t^{2}\)
Explanation: Area $=\frac{1}{2}|t(3t-5t)+2t(5t-2t)+3t(2t-3t)|=\frac{1}{2}|-2t^{2}+6t^{2}-3t^{2}|=\frac{1}{2}|t^{2}|=\frac{t^{2}}{2}$.
Question 4 of 6medium
The line joining A(3,4) and B(7,8) is divided by the point P(a,b) in the ratio 3:5 internally. Find a+b.
10
11
12
13
Explanation: Using the section formula with ratio $3:5$: $a=\frac{3 \times 7+5 \times 3}{8}=\frac{36}{8}=4.5$ and $b=\frac{3 \times 8+5 \times 4}{8}=\frac{44}{8}=5.5$. Hence $a+b=10$.
Question 5 of 6medium
If the midpoint of (2a,4) and (−2,3b) is (1,2a+1), find a+b.
2
3
4
5
Explanation: From earlier example: a=2,b=2→a+b=4
Question 6 of 6medium
Find the reflection of point (4,−3) in the line y = x.
(3,4)
(−3,4)
(3,−4)
(−3,−4)
Explanation: Reflection in the line $y=x$ swaps the coordinates: $(x,y) \to (y,x)$. So $(4,-3)$ maps to $(-3,4)$.
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