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Question 11 Mark
If the distance between the points $(4, k)$ and $(1,0)$ is $5 ,$ then what can be the possible values of $k$ ?
Answer
We know that distance $d$ between two points is given by
$d=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}$
It is given that the distance between the two points is $5$ Putting the values
$\Rightarrow 5=\sqrt{(4-1)^2+(k-0)^2}$
On squaring both sides,
$\Rightarrow 25=9+k^2$
$\Rightarrow 25-9=k^2$
$\Rightarrow 16=k^2$
$\Rightarrow k= \pm 4$
Hence, the possible values of $k$ are $4$ and $-4 .$
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Question 21 Mark
If $\operatorname{ar}(\triangle PQR )$ is zero, then the points $P , Q$ and $R$ are $\qquad$
Answer
collinear
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Question 31 Mark
Find the distance between the points $(a ,b)$ and $(-a,-b)$.
Answer
Image
Distance between two points $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$
$=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(a+a)^2+(b+b)^2}$
$=\sqrt{(2 a)^2+(2 b)^2}$
$=\sqrt{4 a^2+4 b^2}$
$=\sqrt{4\left(a^2+b^2\right)}$
$=2 \sqrt{a^2+b^2}$
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1 Marks Question - Maths STD 10 Questions - Vidyadip