Question

The slope of the straight line passing through A(3, 2) is $\frac{3}{4}$. The coordinates of the points that are 5 units away from A are:

• A. (7, 5)
• B. (-1, -1)
• C. (-7, -5)
• D. (1, 1)

Answer 1

Answer: (A), (B)

(-1, -1)

Let the line makes an angle $\theta$ with X-axis.
$\therefore \tan \theta =\frac{3}{4}\Rightarrow \sin \theta =\frac{3}{5}\mathrm{and}\cos \theta =\frac{4}{5}$
Let B be a point which is at a distance of 5 units from A.
$\therefore$ The equation of line in distance form is:
$\frac{x-3}{\cos \theta }=\frac{y-2}{\sin \theta }\Rightarrow \frac{x-3}{\frac{4}{5}}=\frac{y-2}{\frac{3}{5}}$
and the coordinates of point B will be satisfying the equation
$\frac{x-3}{\frac{4}{5}}=\frac{y-2}{\frac{3}{5}}=\pm 5\mathrm{Let}r=5\mathrm{Then},x-3=\frac{4}{5}\times 5=7\mathrm{and}y-2=\frac{3}{5}\times 5=5$
$\mathrm{Let}r=-5\mathrm{Then},x-3=\frac{4}{5}\times \left(-5\right)=-1\mathrm{and}y-2=\frac{3}{5}\times \left(-5\right)=-1$