The equation of a curve is y= x2-4x+7 and theequation of a line is y+3x = 9. The curve and the line intersect atthe points A and B.?
(i) Find the coordinates of A and of B, and hence, find themidpoint (M) of AB.
ANS: A(2,3) and B(-1,12)
midpointM = (0.5,7.5)
(ii) Find the coordinates of the point Q on the curve at whichthe tangent is parallel to the line y+3x=9.
(iii) Find the distance MQ
Answers
any doubts, comment ;)(
Concept
To find a point in space, use a coordinate, which is like an address. The coordinates of a point in a space with two dimensions are (x, y).
A point's coordinates can be used for a variety of tasks, including calculating distance, midpoint, a line's slope, and its equation.
Given
curve is y = x²₋4x₊7 ....eq(1)
equation of a line is y₊3x = 9 ......eq(2)
Find
we are asked to find out
- the coordinates of A and B
- midpoint of AB
- coordinates of the point Q on the curve at which the tangent is parallel to the line y₊3x = 9
- distance of MQ
Solution
from eq(2) y₊3x = 9 ⇒ y = 9₋3x
now equate eq(2) in eq(1)
9₋3x = x²₋4x₊7
x²₋ 2x ₊ x ₋ 2 = 0
x(x₋2)₊1(x₋2) = 0
(x₊1)(x₋2)=0
x = 1 ; x =2
substitute x values in eq(1)
for x = ₋1
y = 9 ₋ 3x = 9 ₋3(₋1) = 9₊3 = 12
for x = 2
y = 9 ₋ 6 = 3
therefore, intersecting points are
B(x,y) = (₋1,12) and A(2,3)
hence the midpoint of AB = [2₋1/2 , 3₊12/2} = [0.5 , 7.5]
(ii) equation of tangent parallel to y ₊ 3x = 9
slope of tangent m = -3
curve y = y = x²₋4x₊7
differentiating with respect to x
dy/dx = 2x ₋4 = ₋3
2x = 1
x = 1/2
at x = 1/2, y = 1/4-2₊7 = 11/2
therefore Q(x,y) = [1/2,11/2] = [0.5 , 5.5]
Tangent at Q
y ₋ 11/2 = (-3)[x₋1/2]
2y ₋ 11 = ₋6x ₊ 3
6x ₊ 2y ₋ 14 = 0
3x ₊ y ₋ 7 = 0
hence distance between M(0.5 , 7.5) and (0.5 , 5.5) is
MQ = I 7.5 ₋ 5.5 I = 2 units.
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