Grade 10 - Chapter 1 - MCQ Questions

မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။

1. 1. What is the area of the region bounded by the line $y=2-3x$, $x$-axis and $y$-axis?

Explanation

The line $y=2x-3$ cuts x-axis at $\left(\dfrac{2}{3},0\right)$ and y-axis at $(0,2)$. Thus $OX=\dfrac{2}{3}$ and $OY=2$ implies that area of $\triangle XOY=\dfrac{1}{2}\times\dfrac{2}{3}\times2=\dfrac{2}{3}$.

2. 2. If the line $ax+by=-5$ and $x+2y=3$ are perpendicular to each other, then

Explanation

$\begin{array}{l} l_1 : ax+by=-5\\\\ y=-\dfrac{a}{b}x -\dfrac{5}{b}\\\\ m_1=-\dfrac{a}{b}\\\\ l_2 : x+2y=3\\\\ y=-\dfrac{1}{2}x +\dfrac{3}{2}\\\\ m_2=-\dfrac{1}{2}\\\\ l_1\perp l_2\Rightarrow m_1\times m_2=-1\\\\ -\dfrac{a}{b}\left(-\dfrac{1}{2}\right)=-1\\\\ \dfrac{a}{2b}=-1\\\\ a=-2b\\\\ a+2b=0 \end{array}$
3. 3. If $A (4,4), B (8, 7)$ and $C (a,0)$ are such that $AB=AC$, then $a=$.

Explanation

$\begin{array}{l} AB = AC\\\\ \therefore\ AB^2 = AC^2\\\\ (8-4)^2+(7-4)^2=(a-4)^2+(0-4)^2\\\\ 25=(a-4)^2+16\\\\ (a-4)^2=9\\\\ a-4=\pm3\\\\ a=1 \text{ or } 7 \end{array}$
4. 4. The triangle formed by $A (-2, -3), B (1, 3)$ and $C (10,k)$ is a right-angled at $A$, then $k=$.

Explanation

Since $\triangle ABC$ is right-angled at $A$,
$\begin{array}{l} \\ AB\perp AC\\\\ m_{AB}\times m_{AC}=-1\\\\ \dfrac{3-(-3)}{1-(-2)}\times \dfrac{k-(-3)}{10-(-2)}=-1\\\\ \dfrac{6}{3}\times \dfrac{k+3}{12}=-1\\\\ \dfrac{k+3}{6}=-1\\\\ k=-9 \end{array}$
5. 5. Given that $A (3,0), B (4, 5), C (-1, 4)$ and $D (-2,-1)$ are vertices of a rhombus, then the area of $ABCD$ is

Explanation

\begin{aligned} A C &=\sqrt{(-1-3)^{2}+(4-0)^{2}} \\\\ &=\sqrt{32} \\\\ &=4 \sqrt{2} \\\\ B D &=\sqrt{(-2-4)^{2}+(-1-5)^{2}} \\\\ &=\sqrt{72} \\\\ &=6 \sqrt{2} \\\\ & \quad\text { area of } A B C D \\\\ &=\frac{1}{2} \times A C \times B D \\\\ &=\frac{1}{2}(4 \sqrt{2})(6 \sqrt{2}) \\\\ &=24 \end{aligned}
6. 6. The equation of the line through the point $(2, -1)$ and perpendicular to the line $4x-3y=5$ is

Explanation

\begin{aligned} l_{1}\ \text{(given)}: 4 x-3 y &=5 \\\\ y &=\frac{4}{3} x-\frac{5}{3} \\\\ m_{1} &=\frac{4}{3} \\\\ \text { Let } l_{2} & \perp l_{1} . \\\\ \therefore m_{2} &=-\frac{3}{4} \\\\ l_{2}\ \text{(required)}: y-(-1) &=-\frac{3}{4}(x-2) \\\\ 4 y+4 &=-3 x+6 \\\\ 3 x+4 y &=2 \end{aligned}
7. 7. The point which is nearest to the origin is

Explanation

Use distance formula to find the distance of each point from $O(0,0)$
8. 8. What is the perimeter of a triangle with vertices $(1, 4), (1, 7)$ and $(4,4)$ ?

Explanation

Let the vertices be $A(1,4)$, $B(1,7)$ and $C(4,4)$.
\begin{aligned} &\\ A B &=\sqrt{(1-4)^{2}+(7-4)^{2}} \\\\ &=\sqrt{9+9} \\\\ &=3 \sqrt{2} \\\\ B C &=\sqrt{(1-1)^{2}+(4-7)^{2}} \\\\ &=3 \\\\ AC &=\sqrt{(4-1)^{2}+(4-4)^{2}} \\\\ &=3 \\\\ &\quad \text { perimeter }\\\\ &=6+3 \sqrt{2} \end{aligned}
9. 9. If the points $D (-4, 6), E (1, 1)$ and $F(x, y)$ are collinear, then $x + y =$

Explanation

$D(-4,6), E(1,1)$ and $F(x, y)$ are collinear.
\begin{aligned} &\\ m_{D E} &=m_{E F} \\\\ \frac{1-6}{1+4} &=\frac{y-1}{x-1} \\\\ -1 &=\frac{y-1}{x-1} \\\\ 1-x &=y-1 \\\\ x+y &=2 \end{aligned}
10. 10. $\triangle ABC$ is an isosceles right triangle with $\angle B = 90^{\circ}$. If $A= (3, -2), B = (2, -1)$, then $AC=$

Explanation

\begin{aligned} &\\ A C &=\sqrt{2} A B \\\\ &=\sqrt{2} \sqrt{(2-3)^{2}+(-1+2)^{2}} \\\\ &=\sqrt{2} \sqrt{2} \\\\ &=2 \end{aligned}