8 Most FAQ’s of The Straight Line Chapter in Inter 1st Year Maths-1B (TS/AP)

7 Marks

LAQ-1 : Find the circumcentre of the triangle whose vertices are (1,3), (-3,5), (5,-1)

To find the circumcentre S(x, y) of a triangle with vertices A = (1,3) B = (-3,5) and C = (5,-1) follow the steps:

Step 1: Equating $$SA = SB$$

Given SA = SB

$$\sqrt{(x – 1)^2 + (y – 3)^2} = \sqrt{(x + 3)^2 + (y – 5)^2}$$

Squaring both sides we get:

$$(x – 1)^2 + (y – 3)^2 = (x + 3)^2 + (y – 5)^2$$

Expanding and simplifying:

$$(x^2 + 1 – 2x) + (y^2 + 9 – 6y) = (x^2 + 9 + 6x) + (y^2 + 25 – 10y)$$

Further simplification gives:

$$8x – 4y + 24 = 0$$

Which reduces to:

$$2x – y + 6 = 0\quad…(1)$$

Step 2: Equating $$SB = SC$$

Given SB = SC

$$\sqrt{(x + 3)^2 + (y – 5)^2} = \sqrt{(x – 5)^2 + (y + 1)^2}$$

Expanding and simplifying:

$$(x^2 + 9 + 6x) + (y^2 + 25 – 10y) = (x^2 + 25 – 10x) + (y^2 + 1 + 2y)$$

Further simplification gives:

$$16x – 12y + 8 = 0$$

Which reduces to:

$$4x – 3y + 2 = 0\quad…(2)$$

Step 3: Solving Equations (1) and (2)

Equations are:

$$2x – y + 6 = 0$$

$$4x – 3y + 2 = 0$$

Upon solving these equations, the solution provided was:

$$x = -7, y = -10$$

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LAQ-2 : Find the circumcentre of triangle whose vertices are (1,3),(0,-2),(-3,1)

To find the circumcentre S(x, y) of a triangle with vertices A = (1,3) B = (0,-2) and C = (-3,1) we follow the steps you’ve given:

Step 1: Take $$SA = SB$$

$$⇒ \sqrt{(x – 1)^2 + (y – 3)^2} = \sqrt{(x – 0)^2 + (y + 2)^2}$$

Squaring both sides, we get

$$(x – 1)^2 + (y – 3)^2 = (x – 0)^2 + (y + 2)^2$$

$$⇒ (x^2 – 2x + 1) + (y^2 – 6y + 9) = x^2 + (y^2 + 4y + 4)$$

$$⇒ -2x – 10y + 6 = 0$$

$$⇒ x + 5y – 3 = 0…(1)$$

Step 2: Take $$SB = SC$$

$$⇒ \sqrt{(x – 0)^2 + (y + 2)^2} = \sqrt{(x + 3)^2 + (y – 1)^2}$$

Squaring both sides, we get

$$(x – 0)^2 + (y + 2)^2 = (x + 3)^2 + (y – 1)^2$$

$$⇒ x^2 + (y^2 + 4y + 4) = (x^2 + 6x + 9) + (y^2 – 2y + 1)$$

$$⇒ 6x – 6y + 6 = 0$$

$$⇒ x – y + 1 = 0…(2)$$

Step 3: Solving (1) and (2) to get S

From (1), we have

$$x + 5y – 3 = 0$$

From (2), we have

$$x – y + 1 = 0$$

Solving (1) and (2), we obtained the circumcentre S(x, y) as

$$x = -1/3, y = 2/3$$

Thus, the circumcentre S is at $$(-1/3, 2/3)$$

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LAQ-3 : Find the orthocentre of the triangle whose vertices are (5,-2),(-1,2),(1,4)

To find the orthocentre O(x, y) of the triangle with vertices A = (5,-2) B = (-1,2) and C = (1,4) follow the provided steps:

Step 1: Finding Altitude Through A(5,-2)

The slope of BC is $$m = \frac{4 – 2}{1 + 1} = \frac{2}{2} = 1.$$

The perpendicular slope to BC is -1

The equation of the line through A(5,-2) with slope -1 is

$$y – (-2) = -1(x – 5)$$

$$⇒ y + 2 = -x + 5$$

$$⇒ x + y – 3 = 0…(1)$$

Step 2: Finding Altitude Through B(-1,2)

The slope of AC is

$$m = \frac{4 – (-2)}{1 – 5} = \frac{6}{-4} = -\frac{3}{2}.$$

The perpendicular slope to AC is $$\frac{2}{3}$$

The equation of the line through B(-1,2) with slope $$\frac{2}{3}$$

$$y – 2 = \frac{2}{3}(x + 1)$$

$$⇒ 3y – 6 = 2x + 2$$

$$⇒ 2x – 3y + 8 = 0…(2)$$

Step 3: Solving Equations (1) and (2) for O

From (1) we have $$x + y – 3 = 0$$

From (2) we have $$2x – 3y + 8 = 0$$

Solving these equations, we find the orthocentre O(x, y) to be at

$$x = \frac{1}{5}, y = \frac{14}{5}$$

Thus, the orthocentre O(x, y) is $$\left(\frac{1}{5}, \frac{14}{5}\right)$$

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LAQ-4 : Find the orthocentre of the triangle whose vertices are (-2,-1),(6,-1),(2,5)

To find the orthocentre O(x, y) of the triangle with vertices A = (-2,-1) B = (6,-1) and C = (2,5) we follow the steps provided:

Step 1: Finding Altitude Through A(-2,-1)

The slope of BC is $$m = \frac{5 + 1}{2 – 6} = \frac{6}{-4} = -\frac{3}{2}.$$

The perpendicular slope to BC is $$-\frac{1}{m} = \frac{2}{3}.$$

The equation of the line through A(-2,-1) with slope $$\frac{2}{3}$$ is

$$y + 1 = \frac{2}{3}(x + 2),$$

$$3y + 3 = 2x + 4,$$

$$2x – 3y + 1 = 0…(1).$$

Step 2: Finding Altitude Through B(6,-1)

The slope of AC is

$$m = \frac{5 + 1}{2 + 2} = \frac{6}{4} = \frac{3}{2}.$$

The perpendicular slope to AC is $$-\frac{1}{m} = -\frac{2}{3}.$$

The equation of the line through B(6,-1) with slope $$-\frac{2}{3}$$ is

$$y + 1 = -\frac{2}{3}(x – 6),$$

$$3y + 3 = -2x + 12,$$

$$2x + 3y – 9 = 0…(2).$$

Step 3: Solving Equations 1 and 2 for O

From (1) we have $$2x – 3y + 1 = 0$$

From (2) we have $$2x + 3y – 9 = 0$$

Solving these equations, we find:

$$x = 2, y = \frac{5}{3}.$$

Thus, the orthocentre O(x, y) is $$(2, \frac{5}{3})$$

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LAQ-5 : Find the orthocentre of the triangle whose sides are 7x + y – 10 = 0, x – 2y + 5 = 0, x + y + 2 = 0

Given lines are:

$$7x + y – 10 = 0$$

$$x – 2y + 5 = 0$$

$$x + y + 2 = 0$$

Step 1: Finding Altitude Through A

To find A, we solve equations (1) and (2):

$$7x + y – 10 = 0$$

$$x – 2y + 5 = 0$$

Upon solving these, we get A = (1,3)

The slope of line (3) is -1 so its perpendicular slope is 1

The equation of the line through A(1,3) with slope 1 is:

$$y – 3 = 1(x – 1)$$

$$⇒ x – y + 2 = 0…(4)$$

Step 2: Finding Altitude Through B:

To find B we solve equations (1) and (3):

$$7x + y – 10 = 0$$

$$x + y + 2 = 0$$

Upon solving these, we get B = (2,-4)

The slope of line (2) is 1/2 so its perpendicular slope is -2

The equation of the line through B(2,-4) with slope -2 is:

$$y + 4 = -2(x – 2)$$

$$⇒ 2x + y = 0…(5)$$

Step 3: Solving (4) and (5) for O:

Equations (4) and (5) are:

$$2x + y = 0$$

$$x – y + 2 = 0$$

Upon solving these, we find the orthocentre $$O(x, y) = (-2/3, 4/3)$$

Therefore, the orthocentre O(x, y) is $$(-2/3, 4/3)$$

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LAQ-6 : Find the circumcentre of the triangle whose sides are x + y + 2 = 0, 5x – y – 2 = 0, x – 2y + 5 = 0

Given lines:

$$x + y + 2 = 0$$

$$5x – y – 2 = 0$$

$$x – 2y + 5 = 0$$

Step-1: Solving (1) & (2) we get A

From equation (1): $$x + y + 2 = 0$$

From equation (2): $$5x – y – 2 = 0$$

Adding equations (1) and (2):

$$6x = 0$$

$$\Rightarrow x = 0$$

Substituting x = 0 in equation (1):

$$0 + y + 2 = 0$$

$$\Rightarrow y = -2$$

Therefore $$A = (0,-2)$$

Step-2: Solving (1) & (3) we get B;

From equation (1): $$x + y + 2 = 0$$

From equation (3): $$x – 2y + 5 = 0$$

Subtracting equation (3) from equation (1):

$$3y – 3 = 0$$

$$\Rightarrow y = 1$$

Substituting y = 1 in equation (1):

$$x + 1 + 2 = 0$$

$$\Rightarrow x = -3$$

Therefore $$B = (-3,1)$$

Step-3: Solving (2) & (3) we get C;

From equation (3): $$x – 2y + 5 = 0$$

Doubling equation (2): $$10x – 2y – 4 = 0$$

On subtraction:

$$-9x + 9 = 0$$

$$\Rightarrow x = 1$$

Substituting x = 1 in equation (3):

$$1 – 2y + 5 = 0$$

$$\Rightarrow 2y = 6$$

$$\Rightarrow y = 3$$

Therefore $$C = (1,3)$$

Vertices are A = (0,-2), B = (-3,1), C = (1,3).

Step-4: For SA = SB,

The equation simplifies to $$x – y + 1 = 0$$

Step-5: For SB = SC,

The equation simplifies to $$2x + y = 0$$

Step-6: Solving (4) & (5) we get S;

Adding equations (4) and (5):

$$3x + 1 = 0$$

$$\Rightarrow x = -\frac{1}{3}$$

From equation (5):

$$2x + y = 0$$

Substituting $$x = -\frac{1}{3}$$

$$y = -2x$$

$$\Rightarrow y = 2/3$$

Therefore, the circumcenter $$S = (-\frac{1}{3},\frac{2}{3})$$

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LAQ-7 : If Q(h,k) is the foot of the perpendicular of P(x₁,y₁) on the line ax + by + c = 0 then prove that (h – x₁) : n = (k – y₁) : b = -(ax₁ + by₁ + c) : (a² + b²)

Given $$P = (x_1,y_1) Q = (h,k)$$

Slope of PQ is

$$m_1 = \frac{y_2 – y_1}{x_2 – x_1} = \frac{k – y_1}{h – x_1}$$

Slope of the line $$ax + by + c = 0$$ is $$m_2 = – \frac{a}{b}$$

Two lines are perpendicular

$$\Rightarrow m_1 \cdot m_2 = -1$$

$$\Rightarrow \left(\frac{k – y1}{h – x1}\right)\left(- \frac{a}{b}\right) = -1$$

$$\Rightarrow \left(\frac{k – y1}{h – x1}\right)\left(\frac{a}{b}\right) = 1$$

$$\Rightarrow \frac{k – y1}{h – x1} = \frac{b}{a}$$

$$\Rightarrow \frac{k – y1}{b} = \frac{h – x1}{a} = r$$

From (1), we derive:

$$h – x1/a = r$$

$$\Rightarrow h – x1 = ar$$

$$\Rightarrow h = x1 + ar$$

And

$$k – y1/b = r$$

$$\Rightarrow k – y1 = br$$

$$\Rightarrow k = y1 + br$$

But Q(h,k) lies on $$ax + by + c = 0$$

$$\Rightarrow a(h) + b(k) + c = 0$$

$$\Rightarrow a(x1 + ar) + b(y1 + br) + c = 0$$

$$\Rightarrow ax1 + a^2r + by1 + b^2r + c = 0$$

$$\Rightarrow a^2r + b^2r + ax1 + by1 + c = 0$$

$$\Rightarrow r(a^2 + b^2) = -(ax1 + by1 + c)$$

$$\Rightarrow r = -\frac{(ax1 + by1 + c)}{a^2 + b^2}…(2)$$

From (1) & (2)

$$h – x1/a = k – y1/b = -\frac{(ax1 + by1 + c)}{a^2 + b^2}$$

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LAQ-8 : If Q(h,k) is the image of the point P(x₁,y₁) w.r.to the straight line ax + by + c = 0 then prove that (h – x₁) : a = (k – y₁) : b = -2(ax₁ + by₁ + c) : (a² + b²)

Given $$P = (x_1,y_1) Q = (h,k)$$

So, the slope of PQ is:

$$m_1 = \frac{y_2 – y_1}{x_2 – x_1} = \frac{k – y_1}{h – x_1}$$

The slope of the line $$ax + by + c = 0$$ is $$m_2 = – \frac{a}{b}$$

Given that the two lines are perpendicular:

$$m_1 \cdot m_2 = -1$$

$$\Rightarrow \left(\frac{k – y_1}{h – x_1}\right) \cdot \left(- \frac{a}{b}\right) = -1$$

$$\Rightarrow \left(\frac{k – y_1}{h – x_1}\right) \cdot \frac{a}{b} = 1$$

$$\Rightarrow \frac{k – y_1}{h – x_1} = \frac{b}{a}$$

$$\Rightarrow \frac{k – y_1}{b} = \frac{h – x_1}{a}$$

We take $$\frac{h – x_1}{a} = \frac{k – y_1}{b} = r$$

From the equation for r, we have:

$$h – x_1/a = r$$

$$\Rightarrow h – x_1 = ar$$

$$\Rightarrow h = x_1 + ar$$

And

$$k – y_1/b = r$$

$$\Rightarrow k – y_1 = br$$

$$\Rightarrow k = y_1 + br$$

Considering $$P = (x_1, y_1)$$ and $$Q = (h, k)$$

The midpoint of PQ is given by $$R = \left(\frac{x_1 + h}{2}, \frac{y_1 + k}{2}\right)$$

Given that R lies on the line $$ax + by + c = 0$$

$$a\left(\frac{x_1 + h}{2}\right) + b\left(\frac{y_1 + k}{2}\right) + c = 0$$

$$\Rightarrow a(x_1 + h) + b(y_1 + k) + 2c = 0$$

$$\Rightarrow a[x_1 + (x_1 + ar)] + b[y_1 + (y_1 + br)] + 2c = 0$$

$$\Rightarrow a[2x_1 + ar] + b[2y_1 + br] + 2c = 0$$

$$\Rightarrow 2ax_1 + a^2r + 2by_1 + b^2r + 2c = 0$$

$$\Rightarrow r(a^2 + b^2) = -2(ax_1 + by_1 + c)$$

$$\Rightarrow r = -\frac{2(ax_1 + by_1 + c)}{a^2 + b^2}$$

From (1) & (2), we derive:

$$h – \frac{x_1}{a} = k – \frac{y_1}{b} = -\frac{2(ax_1 + by_1 + c)}{a^2 + b^2}$$