Question

Very short answer type questions☆

1. The distance of the point P(2,3) from x-axis is
2. The midpoint of the line segment joining A(3,-7) and (5,-5)
3. The area of a triangle with vertices P(3,0) Q(7,0) and R(8,4) is
_____
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Answer 1

1. The distance between point (2, 3) & x-axis can be determine by assuming a point (2, 0) in x-axis.

Now distance between A & B = P

√ (X2 - X1)² + (Y2 -Y1)²

P = √ (2 - 2)² + (3 - 0)²

P = 3 unit.

3. The area of a ∆PQR with verticles.

P ( X1, Y1) = (3, 0)

Q ( X2, Y2) = (7, 0)

R ( X3, Y3) = (8, 4)

∆ABC = ½ [ X1 (y2 - y3) + X2 (y3 - y1) + X3 (y1 - y2) ]

= ½ {3 (0 - 4) + 7 (4 - 0) + 8 (0-0) } sq.units

= 8 sq.units

Hope it's helps you.

Answer 2

$\underline{\:\large{\sf Answer \: \textit{1} :}}$

As we know that,

$\star\;{\boxed{\sf{\purple{Distance\;formula = \sqrt{( x_2- x_1)^2 + (y_2 - y_1)^2}}}}}$

$:\implies\sf \sqrt{ {(2 - 2)}^{2} + (0 - 3) {}^{2} }$

$:\implies\sf \sqrt{0 + 9}$

$:\implies{\boxed{\frak{\pink{3\;units}}}}\;\bigstar$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{\:\large{\sf Answer \: \textit{2} :}}$

As we know that,

$\star\;{\boxed{\sf{\purple{Mid-point\;formula = \bigg( \dfrac {x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \bigg) }}}}$

$:\implies\sf \bigg( \dfrac {3+5}{2}, \dfrac{-7-5}{2} \bigg)$

$:\implies{\boxed{\frak{\pink{(4,-6)\;}}}}\;\bigstar$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{\:\large{\sf Answer \: \textit{3} :}}$

⠀⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\purple{Area_{ \triangle} = \dfrac{1}{2} \bigg[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg]}}}}$

Therefore,

$:\implies\sf \dfrac{1}{2} \bigg[ 3(0-4)+7(4-0)+8(0-0) \bigg]$

⠀⠀⠀

$:\implies\sf \dfrac{1}{2} \bigg[ -12+8 \bigg]$

$:\implies{\boxed{\frak{\pink{8\;sq.\;units}}}}\;\bigstar$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━