Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 5 Locus and Straight Line Ex 5.4 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.4
Solution & Step-by-Step Answer:
(a) Given equation of the line is 2x + 3y – 6 = 0 Comparing this equation with ax + by + c = 0, we get a = 2, b = 3, c = -6 ∴ Slope of the line = x-intercept = = 3 y-intercept = = 2
(b) Given equation of the line is x + 2y = 0
Comparing this equation with ax + by + c = 0, we get
a = 1, b = 2, c = 0
∴ Slope of the line =
x-intercept = = 0
y-intercept = = 0
Solution & Step-by-Step Answer:
(a) y = 2x – 4 ∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.
(b) y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.
(c)
∴
∴ 2x + y = 4
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.
(d)
∴ 2x = 3y
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
Solution & Step-by-Step Answer:
Let m1 be the slope of the line x – 2y – 7 = 0. ∴ m1 = Let m2 be the slope of the line 2x – 4y + 5 = 0. ∴ m2 = Since, m1 = m2 ∴ The given lines are parallel to each other.
Solution & Step-by-Step Answer:
Let the line 3x + 4y = p cuts the X and Y-axes at points A and B respectively. 3x + 4y = p ∴ ∴ The equation is of the form , with a = and b = ∴ A = (a, 0) = (, 0) and B = (0, b) = (0, ) ∴ OA = and OB = Given, A(∆OAB) = 24 sq. units ∴ × OA × OB = 24 ∴ × × = 24 ∴ p2 = 576 ∴ p = ±24
Solution & Step-by-Step Answer:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC. Let F be the circumcentre of ∆ABC. Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively. ∴ D and E are the midpoints of side BC and AC respectively. ∴ D = = (5, 1) and E = = (1, 3) Now, slope of BC = = -2 ∴ slope of FD = …..[∵ FD ⊥ BC] Since, FD passes through (5, 1) and has slope ∴ Equation of FD is y – 1 = (x – 5) ∴ 2(y – 1) = x – 5 ∴ x – 2y – 3 = 0 ……(i) Since, both the points A and C have same y co-ordinates i.e. 3 ∴ the points A and C lie on the liney = 3. Since, FE passes through E(1, 3). ∴ the equation of FE is x = 1. …….(ii) To find co-ordinates of circumcentre, we have to solve equations (i) and (ii). Substituting the value of x in (i), we get 1 – 2y – 3 = 0 ∴ y = -1 ∴ Co-ordinates of circumcentre F = (1, -1).
Solution & Step-by-Step Answer:
Slope of the line 3x – y + 23 = 0 is 3. ∴ slope of the required line which is perpendicular to 3x – y + 23 = 0 is . Since, the x-intercept of the required line is 3. ∴ it passes through (3, 0). ∴ the equation of the required line is y – 0 = (x – 3) ∴ 3y = -x – 3 ∴ x – 3y = 3
Solution & Step-by-Step Answer:
Let p be the perpendicular distance of the point A(-2, 3) from the line 12x – 5y – 13 = 0 Here, a = 12, b = -5, c = -13, x1 = -2, y1 = 3
Solution & Step-by-Step Answer:
Equations of the given parallel lines are 9x + 6y – 7 = 0 and 9x + 6y – 32 = 0. Here, a = 9, b = 6, C1 = -7 and C2 = -32 ∴ Distance between the parallel lines
Solution & Step-by-Step Answer:
Given equations of lines are x + y – 2 = 0 ……(i) and 2x – 3y – 4 = 0 ……(ii) Multiplying equation (i) by 3, we get 3x – 3y – 6 = 0 …..(iii) Adding equation (ii) and (iii), we get 5x – 2 = 0 ∴ x = Substituting x = in equation (i), we get + y – 2 = 0 ∴ y = 2 – = ∴ The required line passes through point (, ). Also, the line makes intercept of 3 on X-axis ∴ it also passes through point (3, 0). ∴ required equation of line passing through points (, ) and (3, 0) is ∴ 13(5y – 8) = -8(5x – 2) ∴ 65y – 104 = -40x + 16 ∴ 40x + 65y – 120 = 0 ∴ 8x + 13y – 24 = 0 which is the equation of the required line.
Solution & Step-by-Step Answer:
(i) Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ∆ABC. Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ∆ABC. For x-coordinates: Adding (i), (iii) and (v), we get 2x1 + 2x2 + 2x3 = -4 ∴ x1 + x2 + x3 = -2 …..(vii) Solving (i) and (vii), we get x1 = 0 Solving (iii) and (vii), we get x2 = -10 Solving (v) and (vii), we get x3 = 8 For y-coordinates: Adding (ii), (iv) and (vi), we get 2y1 + 2y2 + 2y3 = 6 ∴ y1 + y2 + y3 = 3 …..(viii) Solving (ii) and (viii), we get y1 = -13 Solving (iv) and (viii), we get y2 = 7 Solving (vi) and (viii), we get y3 = 9 ∴ Vertices of ∆ABC are A(0, -13), B(-10, 7), C(8, 9)
(ii) Here, A(0, -13), B(-10, 7), C(8, 9) are the vertices of ∆ABC.
Let F be the circumcentre of ∆ABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC.
∴ D = = (-1, 8)
and E = = (4, -2)
Now, slope of BC =
∴ slope of FD = -9 ……[∵ FD ⊥ BC]
Since, FD passes through (-1, 8) and has slope -9
∴ Equation of FD is y – 8 = -9(x + 1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1 …..(i)
Also, slope of AC =
∴ Slope of FE = ….[∵ FE ⊥ AC]
Since, FE passes through (4, -2) and has slope
∴ Equation of FE is y + 2 = (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = -4x + 16
∴ 4x + 11y = -6 ….(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of y in (ii), we get
∴ 4x + 11(-9x – 1) = -6
∴ 4x – 99x – 11 = -6
∴ -95x = 5
∴ x =
Substituting the value of x in (i), we get
y = -9() – 1 =
∴ Co-ordinates of circumcentre F =
