Question

The slope of a straight line through A(3,2) is 3/4 then the coordinates of the two points on the
line that are 5 units away from A are

Answer 1

Answer :-

→ required points are (7,5) and (-1,-1) .

To Find :-

→ Co-ordinates of two points on the line .

Explanation :-

Given that ;

→ Slope (m) = $\frac{3}{4}$

We know that - distance of a point on the line from a point is "r" which is 5 units .

and, we know about it -

$\sf{from \: points \: (x_{1} , y_{1} )\: } \\ \\ \implies \boxed{ \sf{ \frac{x -x_{1} }{ \cos \theta} = \frac{y -y_{1} }{ \sin \theta} = \pm \: r \: }}$

We have point A( 3,2) and slope (m)

$\because \sf{ slope \: (m) = \tan \theta \: } \\ \\ \to \: \frac{3}{4} = \tan \theta \: \\ \\ \implies \boxed{ \theta = 37 \degree} \\ \\ \therefore \sf{ \sin 37 \degree = \frac{3}{5} \: and \: \cos 37 \degree = \frac{4}{5} } \\ \\ \bf{ hence} \\ \\ \to \sf{ \frac{x - 3}{ \frac{4}{5} } = \frac{y - 2}{ \frac{3}{5} } = \pm 5} \\ \\ (1) \sf{for \: x \: \: co-ordinate \: } \\ \\ \to \sf{ \frac{x - 3}{ \frac{4}{5} } = \pm5 \: }$

$\to \sf{x - 3 = \pm \: 5 \times \frac{4}{5} } \\ \\ \to \sf{x - 3 = \pm 4} \\ \\ \therefore \boxed{ \sf{ \: x = 7 \:, \: -1}} \\ \\ (2) \sf{for \: y \: co - ordinate} \\ \\ \to \sf{ \frac{y - 2}{ \frac{3}{5} } = \pm5} \\ \\ \to \sf{ y - 2 = \pm 3} \\ \\ \to \boxed{\sf{y = 5 \: , \: - 1}}$

hence required points are (7,5) and (-1,-1)

Answer 2

$\tt{\huge{\green{Solution \: :- }}}$

QUESTION :

The slope of a straight line through A(3,2) is 3/4 then the coordinates of the two points on the line that are 5 units away from A are _______

SOLUTION :

Slope, m = 3/4

Answer :-

→ required points are (7,5) and (-1,-1) .

To Find :-

→ Co-ordinates of two points on the line .

Explanation :-

Given that ;

→ Slope (m) = \frac{3}{4}

4

3

Distance of a point on the line from a point is "r" which is 5 units .

$\begin{lgathered}\sf{from \: points \: (x_{1} , y_{1} )\: } \\ \\ \implies \boxed{ \sf{ \frac{x -x_{1} }{ \cos \theta} = \frac{y -y_{1} }{ \sin \theta} = \pm \: r \: }}\end{lgathered}$

$\begin{lgathered}\because \sf{ slope \: (m) = \tan \theta \: } \\ \\ \to \: \frac{3}{4} = \tan \theta \: \\ \\ \implies \boxed{ \theta = 37 \degree} \\ \\ \therefore \sf{ \sin 37 \degree = \frac{3}{5} \: and \: \cos 37 \degree = \frac{4}{5} } \\ \\ \bf{ hence} \\ \\ \to \sf{ \frac{x - 3}{ \frac{4}{5} } = \frac{y - 2}{ \frac{3}{5} } = \pm 5} \\ \\ (1) \sf{for \: x \: \: co-ordinate \: } \\ \\ \to \sf{ \frac{x - 3}{ \frac{4}{5} } = \pm5 \: }\end{lgathered}$

$\begin{lgathered}\to \sf{x - 3 = \pm \: 5 \times \frac{4}{5} } \\ \\ \to \sf{x - 3 = \pm 4} \\ \\ \therefore \boxed{ \sf{ \: x = 7 \:, \: -1}} \\ \\ (2) \sf{for \: y \: co - ordinate} \\ \\ \to \sf{ \frac{y - 2}{ \frac{3}{5} } = \pm5} \\ \\ \to \sf{ y - 2 = \pm 3} \\ \\ \to \boxed{\sf{y = 5 \: , \: - 1}}\end{lgathered}$