17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

 17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

 

B (4, –1) and C (1, 2)

using two point form equation of BC is 

{(y-(-1)) / [2-(-1)]} = {(x-4) / (1-4)} 

[y+1] / 3 = [x-4] / (-3)

y+1 = -x+4

x+y-3 = 0--------------(1)

 

line perpendicular to ax+by+c=0 is of the form bx-ay+k=0

 

altitude through A is perpendicular to BC

so the form of the altitude is

x-y+k=0--------------------(2)

This passes through A(2,3)

so 

2-3+k =0

k = 1

so required altitude through A is

x - y +1 =0 using k = 1 in (2)

solving  equation of altitude with that of BC (1)

 

x -  y = (-1)

x + y = 3

solving 

2x = 2

x = 1

resubstitute

y =2

point of intersection is P(1,2)

A is (2,3)

 

using distance formula

length of the altitude is

sqrt[ {(2-1)^2} + {(3-2)^2}] 

=sqrt[2]

we can also use perpendicular distance formula to find the distance between

 A(2,3)  and the line BC { x + y = 3 } for finding the length of the altitude.

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

Given line is 3x – 4y – 16 = 0

line perpendicular to ax+by+c=0 is of the form bx-ay+k=0

line perpendicular to given line is

-4x-3y+k = 0

This passes through  (–1, 3) if

-4(-1) - 3(3) + k =0

k =5

so perpendicular line is

-4x-3y+5 = 0

or 4x + 3y  = 5

given equation is

3x - 4y = 16

solving

[4x + 3y  = 5  ] *3

[3x - 4y = 16  ]  *4

 

12x+9y =15

12x-16y=64

 

subtracting

25y =(-49) 

y = [(-49) / 25]

resubstitute and solve for x

3x -4 [(-49) / 25] = 16

3x =16 -[196/25]

3x =204/25

x =68/25

 

foot of the perpendicular is [(68/25) , (-49/25)]

 

 

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ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 

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