RD Chapter 23 The Straight Lines Ex 23.10 Solutions
Question - 1 : - Find the point of intersection of the following pairs of lines:
(i) 2x – y + 3 = 0 and x + y – 5 = 0
(ii) bx + ay = ab and ax + by = ab
Answer - 1 : -
(i) 2x – y + 3 = 0 and x +y – 5 = 0
Given:
The equations of thelines are as follows:
2x − y + 3 = 0 … (1)
x + y − 5 = 0 … (2)
Let us find the pointof intersection of pair of lines.
By solving (1) and (2)using cross – multiplication method, we get
x = 2/3 and y = 13/3
∴ The point ofintersection is (2/3, 13/3)
(ii) bx + ay = ab and ax +by = ab
Given:
The equations of thelines are as follows:
bx + ay − ab = 0… (1)
ax + by = ab ⇒ ax + by − ab = 0 … (2)
Let us find the pointof intersection of pair of lines.
By solving (1) and (2)using cross – multiplication method, we get
∴ The point ofintersection is (ab/a+b, ab/a+b)
Question - 2 : - Find the coordinates of the vertices of a triangle, the equations ofwhose sides are: (i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0
(ii) y (t1 + t2) = 2x + 2at1t2,y (t2 + t3) = 2x + 2at2t3 and,y(t3 + t1) = 2x + 2at1t3.
Answer - 2 : -
(i) x + y – 4 = 0, 2x – y+ 3 0 and x – 3y + 2 = 0
Given:
x + y − 4 = 0, 2x − y+ 3 = 0 and x − 3y + 2 = 0
Let us find the pointof intersection of pair of lines.
x + y − 4 = 0 … (1)
2x − y + 3 = 0 … (2)
x − 3y + 2 = 0 … (3)
By solving (1) and (2)using cross – multiplication method, we get
x = 1/3, y = 11/3
Solving (1) and (3)using cross – multiplication method, we get
x = 5/2, y = 3/2
Similarly, solving (2)and (3) using cross – multiplication method, we get
x = – 7/5, y = 1/5
∴ The coordinatesof the vertices of the triangle are (1/3, 11/3), (5/2, 3/2) and (-7/5, 1/5)
(ii) y (t1 +t2) = 2x + 2at1t2, y (t2 + t3)= 2x + 2at2t3 and, y(t3 + t1)= 2x + 2at1t3.
Given:
y (t1 +t2) = 2x + 2a t1t2, y (t2 + t3)= 2x + 2a t2t3 and y (t3 + t1)= 2x + 2a t1t3
Let us find the pointof intersection of pair of lines.
2x − y (t1 +t2) + 2a t1t2 = 0 … (1)
2x − y (t2 +t3) + 2a t2t3 = 0 … (2)
2x − y (t3 +t1) + 2a t1t3 = 0 … (3)
By solving (1) and (2)using cross – multiplication method, we get
Solving (1) and (3)using cross – multiplication method, we get
Solving (2) and (3)using cross – multiplication method, we get
∴ The coordinatesof the vertices of the triangle are (at21, 2at1), (at22,2at2) and (at23, 2at3).
Question - 3 : - Find the area of the triangle formed by the lines
y = m1x + c1, y = m2x + c2 andx = 0
Answer - 3 : -
Given:
y = m1x + c1 …(1)
y = m2x + c2 …(2)
x = 0 … (3)
In triangle ABC, letequations (1), (2) and (3) represent the sides AB, BC and CA, respectively.
Solving (1) and (2),we get
Solving (1) and (3):
x = 0, y = c1
Thus, AB and CAintersect at A 0,c1.
Similarly, solving (2)and (3):
x = 0, y = c2
Thus, BC and CAintersect at C 0,c2.
Question - 4 : - Find the equations of the medians of a triangle, the equations of whosesides are:
3x + 2y + 6 = 0, 2x – 5y + 4 = 0 and x – 3y – 6 = 0
Answer - 4 : -
Given:
3x + 2y + 6 = 0 … (1)
2x − 5y + 4 = 0 … (2)
x − 3y − 6 = 0 … (3)
Let us assume, intriangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA,respectively.
Solving equations (1)and (2), we get
x = −2, y = 0
Thus, AB and BCintersect at B (−2, 0).
Now, solving (1) and(3), we get
x = – 6/11, y = –24/11
Thus, AB and CAintersect at A (-6/11, -24/11)
Similarly, solving (2)and (3), we get
x = −42, y = −16
Thus, BC and CAintersect at C (−42, −16).
Now, let D, E and F bethe midpoints the sides BC, CA and AB, respectively.
Then, we have:
∴ The equations ofthe medians of a triangle are: 41x – 112y – 70 = 0,
16x – 59y – 120 = 0,25x – 53y + 50 = 0
Question - 5 : - Prove that the lines y = √3x + 1, y = 4 and y = -√3x + 2 form an equilateral triangle.
Answer - 5 : -
Given:
y = √3x + 1…… (1)
y = 4 ……. (2)
y = – √3x + 2……. (3)
Let us assume intriangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA,respectively.
By solving equations(1) and (2), we get
x = √3, y = 4
Thus, AB and BCintersect at B(√3,4)
Now, solving equations(1) and (3), we get
x = 1/2√3, y = 3/2
Thus, AB and CAintersect at A (1/2√3, 3/2)
Similarly, solvingequations (2) and (3), we get
x = -2/√3, y = 4
Thus, BC and ACintersect at C (-2/√3,4)
Now, we have:
Hence proved, thegiven lines form an equilateral triangle.