Question

find the coordinates of the foot of perpendicular drawn from the point A(-2,3) to the line 3x-y-1=o​

Answer 1

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Let ,

C(a,b) be the coordinate of foot perpendicular

Given ,

A(-2,3) is the point on the line BC

The equation of line ED is 3x - y - 1 = 0

Thus ,

$\sf \mapsto 3x - y - 1 = 0 \\ \\ \sf \mapsto y = 3x - 1$

This equation is of the form y = mx + c , where m = 3 and c = -1

Hence , the slope of ED is 3

Since , the lines BC and ED are perpendicular to each other

So , their product of slopes is -1

Thus ,

$\sf \mapsto 3 \times m_{2} = - 1 \\ \\\sf \mapsto m_{2} = - \frac{1}{3}$

Hence , the slope of BC is -1/3

The line BC passes through the point A(-2,3) and C(a,b)

We know that , point slope form is given by

$\large { \star}$ m = y2 - y1/x2 - x1

So,

$\sf \mapsto - \frac{1}{3} = \frac{b - 3}{a - ( - 2)} \\ \\\sf \mapsto - \frac{1}{3} = \frac{b - 3}{a + 2} \\ \\ \sf \mapsto - a - 2 = 3b - 9 \\ \\\sf \mapsto \fbox{a + 3b - 7 = 0}$

Also , C(a,b) lies on the line ED

So , point C(a,b) satisfy the equation of the line ED 3x - y - 1 = 0

Putting x = a and y = b , we obtain

$\sf \mapsto \fbox{3a - b - 1 = 0}$

Now , we get our equations which are

$\sf \mapsto a + 3b = 7 ---(i) \\ \\ \sf \mapsto 3a - b = 1--- (ii)$

Multiply eq (i) by 3 and eq (ii) by 1 , we get

$\sf \mapsto 3a + 9b = 21--- (iii) \\ \\ \sf \mapsto 3a - b = 1--- (iv)$

Subtract eq (iv) from eq (iii) , we get

$\sf \mapsto 10b = 20 \\ \\ \sf \mapsto b = \frac{20}{10} \\ \\ \sf \mapsto b = 2$

Put the value of b = 2 in eq (i) , we get

$\sf \mapsto a + 3(2) = 7 \\ \\ \sf \mapsto a + 6 = 7 \\ \\\sf \mapsto a = 7 - 6 \\ \\ \sf \mapsto a = 1$

Hence , the C(1,2) is coordinates the of the foot perpendicular

Answer 2

C(a,b) be the coordinate of foot perpendicular

Given ,

A(-2,3) is the point on the line BC

The equation of line ED is 3x - y - 1 = 0

Thus ,

This equation is of the form y = mx + c , where m = 3 and c = -1

Hence , the slope of ED is 3

Since , the lines BC and ED are perpendicular to each other

So , their product of slopes is -1

Thus ,

Hence , the slope of BC is -1/3

The line BC passes through the point A(-2,3) and C(a,b)

We know that , point slope form is given by

m = y2 - y1/x2 - x1

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