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Beranda / algebra / chapter-1 / coordinate geometry / grade 10 / mcq / multiple choice / new syllabus

Grade 10 - Chapter 1 - MCQ Questions

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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






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










    11. Your Score:





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