Find the equation of the striaght lines passing through the following pair of points:
(i) (0, 0) and (2, - 2) (ii) (a, b) and ( a + c sin α, b + c cos α)
(iii) (0, - a) and (b, 0) (iv) (a, b) and (a + b, a - b) (v) (at1, a/t1) and (at2, a/t2) (vi) (a cos α, a sin α) and (a cos β, a sin β)
Find the equation of the striaght lines passing through the following pair of points:
(i) (0, 0) and (2, - 2) (ii) (a, b) and ( a + c sin α, b + c cos α)
(iii) (0, - a) and (b, 0) (iv) (a, b) and (a + b, a - b) (v) (at1, a/t1) and (at2, a/t2) (vi) (a cos α, a sin α) and (a cos β, a sin β)
(i) Here,(x1y1)=(0, 0)(x2y2)=(2,−2)The equation of the given straight line is :y−y1=m (x−x1)⇒ y−y1=y2−y1x2−x1 (x−x1)⇒ y−0=−2−02−0(x−0)⇒ y=−2x2⇒ y=−x∴ The equation of the line joining the points(0, 0) and (2,−2) is y=−x(ii) Let A (a, b)=(x1 y1)B(a + c sin a, b + c cos α)=(x2y2)Then equation of line AB is⇒ y−y1=y2−y1x2−x1 (x−x1)⇒ y−b=b+c cos α−ba+c sin α−a(x−a)⇒ y−b=c cot αc sin α(x−a)⇒ y−b=cot α (x−a)∴ The equation of the line joining the points(a, b) and (a + c sin α, b + c cos α) isy−b=cot α (x−a)(iii) Let A(0, −a) be (x1y1)B(b, 0) be (x2y2)Then equation of line AB is⇒ y−y1=y2−y1x2−x1 (x−x1)⇒ y−(−a)=0−(−a)b−0(x−0)⇒y+a=ab(x−0)⇒ax−by=ab∴ The equation of the line joining the points(0, −a) and (b, 0) is ax−by=ab.(iv) Let A(a, b) be (x1y1)B(a+b, a−b) be (x2y2)Then equation of line AB is⇒ y−y1=y2−y1x2−x1(x−x1)⇒ y−b=a−b−ba+b−a(x−a)⇒ y−b=a−2bb(x−a)⇒ by−b2=ax−a2−2bx+2ba⇒ (a−2b)x−by+b2−a2+2ab=0∴ The equation of the line joining the points(a, b) and (a+b, a−b) is (a−2b)x−by+b2−a2+2ab=0(v) Let A(x1y1) be (at1,at1)B(x2y2) be (at2,at2)Then equation of line AB is⇒y−y1=y2−y1x2−x1 (x−x1)⇒y−at1=at2−at1at2−at1(x−at1)⇒y−at1=a(t2−t2)at1t2(t2−t1)(x−at1)⇒y−at1=−1t1t2(x−at1)⇒ t1t2y+x=a(t1+t2)∴ The equation of the line joining the points(at1,at1) and (at2,at2) is t1t2y+x=a(t1+t2)(vi) Let A(x1y1) be (a cos α, a sin α)B(x2y2) be (a cos β, a sin β)⇒y−y1=y2−y1x2−x1 (x−x1)⇒y−a sin α=asinβ−asinαacoβ−acosα(x−acosα)⇒y−a sin α=a(−2sin(β−α2))cosβ(β+α2)a(−2sin(β−α2))sin(β+α2)(x−acosα)⇒y−a sin α=cos(α+β2)sin(β+α2)(x−acosα)⇒x cos(α+β2)+y sin α+β2=a cosα−β2∴ The equation of the line joining the points(a cos α, a sin α) and (a cos β, a sin β)is x cos (α+β2)+y sin (α+β2)=a cos α−β2
(i) Here,(x1y1)=(0, 0)(x2y2)=(2,−2)The equation of the given straight line is :y−y1=m (x−x1)⇒ y−y1=y2−y1x2−x1 (x−x1)⇒ y−0=−2−02−0(x−0)⇒ y=−2x2⇒ y=−x∴ The equation of the line joining the points(0, 0) and (2,−2) is y=−x(ii) Let A (a, b)=(x1 y1)B(a + c sin a, b + c cos α)=(x2y2)Then equation of line AB is⇒ y−y1=y2−y1x2−x1 (x−x1)⇒ y−b=b+c cos α−ba+c sin α−a(x−a)⇒ y−b=c cot αc sin α(x−a)⇒ y−b=cot α (x−a)∴ The equation of the line joining the points(a, b) and (a + c sin α, b + c cos α) isy−b=cot α (x−a)(iii) Let A(0, −a) be (x1y1)B(b, 0) be (x2y2)Then equation of line AB is⇒ y−y1=y2−y1x2−x1 (x−x1)⇒ y−(−a)=0−(−a)b−0(x−0)⇒y+a=ab(x−0)⇒ax−by=ab∴ The equation of the line joining the points(0, −a) and (b, 0) is ax−by=ab.(iv) Let A(a, b) be (x1y1)B(a+b, a−b) be (x2y2)Then equation of line AB is⇒ y−y1=y2−y1x2−x1(x−x1)⇒ y−b=a−b−ba+b−a(x−a)⇒ y−b=a−2bb(x−a)⇒ by−b2=ax−a2−2bx+2ba⇒ (a−2b)x−by+b2−a2+2ab=0∴ The equation of the line joining the points(a, b) and (a+b, a−b) is (a−2b)x−by+b2−a2+2ab=0(v) Let A(x1y1) be (at1,at1)B(x2y2) be (at2,at2)Then equation of line AB is⇒y−y1=y2−y1x2−x1 (x−x1)⇒y−at1=at2−at1at2−at1(x−at1)⇒y−at1=a(t2−t2)at1t2(t2−t1)(x−at1)⇒y−at1=−1t1t2(x−at1)⇒ t1t2y+x=a(t1+t2)∴ The equation of the line joining the points(at1,at1) and (at2,at2) is t1t2y+x=a(t1+t2)(vi) Let A(x1y1) be (a cos α, a sin α)B(x2y2) be (a cos β, a sin β)⇒y−y1=y2−y1x2−x1 (x−x1)⇒y−a sin α=asinβ−asinαacoβ−acosα(x−acosα)⇒y−a sin α=a(−2sin(β−α2))cosβ(β+α2)a(−2sin(β−α2))sin(β+α2)(x−acosα)⇒y−a sin α=cos(α+β2)sin(β+α2)(x−acosα)⇒x cos(α+β2)+y sin α+β2=a cosα−β2∴ The equation of the line joining the points(a cos α, a sin α) and (a cos β, a sin β)is x cos (α+β2)+y sin (α+β2)=a cos α−β2
