Question

$\huge\green{\boxed{\sf Question}}$

If the slope of a line passing through the point A(3, 2) is
3/4, then find
points on the line which are 5 units away from the point A​

Answer 1

$\huge\green{\boxed{\sf Question}}$

If the slope of a line passing through the point A(3, 2) is

3/4, then find

points on the line which are 5 units away from the point A

$\huge{answer}$

Equation of line with given slope  43 and from point A(3,2)

Equation of line with given slope  43 and from point A(3,2)y−2=43(x−3)

Equation of line with given slope  43 and from point A(3,2)y−2=43(x−3)4y−8=3x−9

Equation of line with given slope  43 and from point A(3,2)y−2=43(x−3)4y−8=3x−93x=1+4y------(1)

Let Point B(x1,y1)

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula(x1−3)2+(y1−2)2=25

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula(x1−3)2+(y1−2)2=25x12+9−6x1+y12+4−4y1=25

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula(x1−3)2+(y1−2)2=25x12+9−6x1+y12+4−4y1=25x12+y12−6x1−4y1−12=

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula(x1−3)2+(y1−2)2=25x12+9−6x1+y12+4−4y1=25x12+y12−6x1−4y1−12=Point B lies on line 3x=1+4y

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula(x1−3)2+(y1−2)2=25x12+9−6x1+y12+4−4y1=25x12+y12−6x1−4y1−12=Point B lies on line 3x=1+4yHence x1=34y1+1

Let Point B(x1,y1)The distance between A and B is 5 Hence by distance formula(x1−3)2+(y1−2)2=25x12+9−6x1+y12+4−4y1=25x12+y12−6x1−4y1−12=Point B lies on line 3x=1+4yHence x1=34y1+1(34y1+1)

Answer 2

Answer:

$\huge \red{AnsWEr }$

Let the coordinates of A be (3,2).

Let P be a point at a distance of 5 units from A. AP has a slope of 3/4.

Draw triangle APM, where M is the foot of the perpendiculars from A and P drawn parallel to the x- and y-axes.

Triangle APM has AM = 4 and PM = 3, as APM is a RAT. Coordinates of M = [(3+4),2)] or (7,2) and coordinates of P are [7,(2+3)] or (7,5).

Distance AP =$[(7 - 3) {}^{2} + (5 - 2) {}^{2} ] {}^{0.5}$

$= [4 {}^{2} + 3 {}^{2}] {}^{0.5} \\ = [16 + 9] {}^{0.5} \\ = 25 {}^{0.5} \\ = 5$

Coordinates Of P = (7,5)