10th GRADE MATH SOLUTION EXERCISE 5.4 PART8 10th grade math solution exercise 5.4 part8 :

Here we are going to see solutions of some problems that we find in the exercise number 5.4

Question 18 :

Find the equation of the line passing through (22, -6) and having intercept on x-axis exceeds the intercept on y-axis by 5.

Solution :

x-intercept(a) = b + 5, y -intercept = b

(x/a) + (y/b)  =  1

x/(b+5) + y/b  =  1

The straight line is passing through the point (22, -6)

22/(b+5) - 6/b  =  1

22b - 6(b + 5)  =  b(b + 5)

22b - 6b - 30  =  b² + 5b

16b - 30 = b² + 5b

b² + 5b - 16b + 30 = 0

b² - 11b + 30 = 0

(b - 5) (b - 6) = 0

 b - 5 = 0 ==> b = 5a = 5 + 5 = 10 ==> a = 10x/a + y/b = 1x/10 + y/5 = 1(x + 2y)/10 = 1x - 2y = 10x + 2y - 10 = 0 b - 6 = 0 ==> b = 6a = 6 + 5 = 11 ==> a = 11x/a + y/b = 1x/11 + y/6 = 1(6x + 11y)/66 = 16x + 11y = 666x + 11y - 66 = 0

Question 19 :

If A (3, 6) and C (-1, 2) are two vertices of a rhombus ABCD, then find the equation of straight line that lies along the diagonal BD.

Solution : In any rhombus diagonals bisect each other at right angle.

In any rhombus midpoint of the diagonals will be equal.

Midpoint of AC  =  Midpoint of BD

Midpoint of AC = (x₁+x₂)/2, (y₁+y₂)/2

=  (3 + (-1))/2, (6 + 2)/2

=  2/2, 8/2

=  (1, 4)

(1, 4) is a point lies of the diagonal BD.

Slope of AC x Slope of BD = -1

Slope of AC :

m  =  (y₂-y₁)/(x₂-x₁) ==> (2-6)/(-1-3) ==> -4/(-4)  ==> 1

Slope of BD :

Slope of BD = -1/1 ==> -1

Equation of BD :

(y - y₁)  =  m (x - x₁)

y - 4 = -1(x - 1)

y - 4 = -x + 1

x + y - 4 - 1 = 0

x + y - 5 = 0

Hence the required equation is x + y - 5 = 0.

Question 20 :

Find the equation of the line whose gradient is 3/2 and which passes through P, where P divides the line segment joining A(-2, 6) and B (3, -4) in the ratio 2 : 3 internally.

Solution :

First we need to find the point P.

P = (lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m)

p =  (2(3) + 3(-2))/(2+3), (2(-4) + 3(6))/(2+3)

=  (6 -6)/5, (-8 + 18)/5

=  0/5, 10/5

=  (0, 2)

(x₁, y₁) ==> (0, 2) m = 3/2

(y - y₁) = m(x - x₁)

(y - 2) = (3/2)(x - 0)

2(y - 2) = 3x

2y - 4 = 3x

3x - 2y + 4 = 0

Hence the required equation is 3x - 2y + 4 = 0.

After having gone through the stuff given above, we hope that the students would have understood "10th grade math solution exercise 5.4 part8".

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