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Correct option is (A) \(\frac{x}{a} + \frac{y}{b} = 1\)

\(\frac{x}{a} + \frac{y}{b} = 1\) is intercept form a line.

Correct option is A) \(\cfrac{x}{a}\) + \(\cfrac{y}{b}\) = 1

Comparing y = 5\(x\) –7 with y = m\(x\) + c, the slope of given line = m = 5 

∴ Equation of a line parallel to y = 5\(x\) – 7 having y-intercept = –1 is y = 5\(x\) – 1.

Since P (x, y) is equidistant from A (5, 1) and B (-1, 5), then

PA = PB

or PA² = PB²

(5 – x)² + (1 – y)² = (– 1 – x)² + (5 – y)²

(25 + x² – 10x) + (1 + y² – 2y) = (1 + x² + 2x + 25 + y² – 10y)

26 + x² – 10x + y² – 2y = (26 + x² + 2x + y² – 10y)

12x = 8y

3x = 2y

Hence proved.

Since P (x, y) is equidistant from A(6, -1) and B(2, 3), then

PA = PB

or PA² = PB²

(6 – x)² + (-1 – y)² = (2 – x)² + (3 – y)²

(36 + x² – 12x) + (1 + y² + 2y) = (4 + x² – 4x + 9 + y² – 6y)

37 – 12x + 2y = 13 – 4x – 6y

8x = 8y + 24

x – y = 3

Hence proved.

(c) \(\frac{b+k}{f+h}\)

Let the slope of the lin passing through the points (–k, h) and (b, – f) be m₁. Then m₁ = \(\frac{-f-h}{b+k}\) = \(-\bigg(\frac{f+h}{b+k}\bigg)\)

\(\bigg[Slope = \frac{y_2-y_1}{x_2-x_1}\bigg]\)

∴ Slope of line perpendicular to the given line = \(-\frac{1}{m_1}\)

= \(\frac{-1}{-\big(\frac{f+h}{b+k}\big)}\) = \(\bigg(\frac{b+k}{f+h}\bigg)\)

If x ≤ 0 or x ≥ 1 then x² ≥ x. 

For every integer x, x² ≥ x.

The points (ii) (0, 0), (iii) (3, 0) (iv) (- 5, 0) (vi) (- 6, 0) and (ix) (b, 0) lie on X – axis.

A(- 3, 4) ; B(0, 5) : C (3, 4) ; D (2, 4) ; E (2, 0) ; 

F (3, 0) ; G (3, – 1) ; H (0, – 1) ; I (- 3, – 1) ; 

J (- 3, 0) ; K (- 2, 0) ; L (- 2, 4) ; M (- 1, 0) ; 

N (-1, 3); O (0, 0) ; P (1. 3) and Q (1, 0)

Given point = M(4, -3)

Distance from x -axis to the M is | -3 | units = 3 

Distance from y – axis to the M is | 4 | units = 4

i) A(2, 9) ; B(5, 9); C(2, 6) ; D(5, 3) ; E(2, 3) 

ii) F(- 6, – 2) ; G(- 4, – 5) ; H(- 3, – 7) ; I(- 9, – 7) ; J(- 8, – 5)

i. Distance of line LM from the Y-axis is 3 units. 

ii. P(3, 2), Q (3, -1), R(3, 0) 

iii. x co-ordinate of point L = 3 

x co-ordinate of point M = 3 

∴ Difference between the x co-ordinates of the points L and M = 3 – 3 

= 0

Quadrant Position

(i) In point (-3, 5), x-coordinate is negative and y-coordinate is positive, so it lies in
II quadrant.

(ii) In point (4,-1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.

(iii) In point (2,0), x-coordinate is positive and y-coordinate is zero, so it lies on X-axis.

(iv) In point (2,2), x-coordinate and y-coordinate both are positive, so it lies in I quadrant.

(v) In point (-3, – 6), x-coordinate and y-coordinate both are negative, so it lies in III quadrant.

(i) The point is (-3,5).

Hence, the point lies in the II quadrant.

(ii) The point is (-5,-3).

Hence, the point lies in the III quadrant.

(iii) The point is (-5,3).

Hence, the point lies in the II quadrant.

(iv) The point is (3,5).

Hence, the point lies in the I quadrant.

Coordinate of P (5, 3)

Coordinate of Q (-4, 6)

Coordinate of R (-3, -2)

Coordinate of S = (1, -5)

Correct answer is

(D) (2, 0)

The correct answer is : (D) (2, 0)

Let the coordinates of Point Q are (x, y) and coordinates of origin O are (0, 0) 

Since OP = OQ 

Points are: P (-3, 2), Q(x, y) and O (0, 0) 

Q is the mid point

0 = \(\frac{-3 + x}2\)

x = 3

0 = \(\frac{2 + y}2\)

y = - 2

Therefore coordinates are (3, -2)

The positions of (5, -8) and (-8, 5) are not same. 

They are two distinct points. (5, -8) lies at a distance of 5 units from Y – axis and 8 units from X – axis on down side of the origin. So it lies in Q₄ . Where as (- 8, 5) lies in Q₂. The point is at a distance of 8 units from Y – axis on left side of the origin and 5 units from X – axis.

i. Quadrant I

ii. Quadrant III

iii. Quadrant IV 

iv. Quadrant II

Correct option is (A) (a, a)

Correct option is (B) y = 0

i. Q(-2, 2) and R(4, -1) 

ii. T(0, -1) and M(3, 0) 

iii. Point S lies in the third quadrant. 

iv. The x and y co-ordinates of point O are equal.

(C) The answer is third

(B) The answer is Q and R

Correct option is (A) x > 0, y < 0

If (x, y) lies in \(4^{th}\) quadrant.

Then x > 0 & y < 0

A) x > 0, y < 0

The y co-ordinate of all the points is the same. 

∴ The line which passes through the given points is parallel to X-axis.

 Correct option is (C) Parallel to X-axis

i. The equation of a line parallel to the Y-axis is x = a. 

∴ The line x = 3 is parallel to the Y-axis. 

ii. y – 2 = 0 ∴ y = 2 The equation of a line parallel to the X-axis is y = b. 

∴ The line y – 2 = 0 is parallel to the X-axis.

iii. x + 6 = 0 ∴ x = -6 The equation of a line parallel to the Y-axis is x = a. 

∴ The line x + 6 = 0 is parallel to the Y-axis. 

iv. The equation of a line parallel to the X-axis is y = b. 

∴ The line y = – 5 is parallel to the X-axis

(C) Parallel to X-axis

The y co-ordinate of all the points is the same. 

∴ The line which passes through the given points is parallel to X-axis.

No, the given points (5, 7) and (7, 5) are not equal. 

Why because, the above two points represent two different points in the coordinate plane. 

So they are not equal.

For (x₁, y₁) = (x₂, y₁) then x₁ must be equal to x₂ and y₁ must be equal to y₂. 

So to become (5, 7) = (7, 5) 5 should be equal to 7, which is impossible. 

So (5, 7) will not be equal to (7, 5).

C. on x – axis

Explanation:

(-5,2) is of the form (-x,y) so it lies in the II quadrant.

(2,-5) is of the form (x,-y) so it lies in IV quadrant.

(C) II and IV quadrants, respectively

Hence, (C) is the correct option.

D. y – coordinate = 5 or –5

Explanation:

Perpendicular distance from x-axis = Ordinate = 5

The negative direction of x-axis doesn’t decide the sign of the ordinate.

(D) y-coordinate = 5 or -5.

Hence, (D) is the correct option.

(b) 1

We have, points P(- 2, 3) and Q(- 3, 5)

Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = – 2 – (-3) = -2 + 3 = 1.

Correct option is (B) OX’

y-coordinate of point (-2, 0) is O.

\(\therefore\) (- 2, 0) lies on negative x-axis.

i.e., (- 2, 0) lies on OX'.

Correct option is  B) OX’

Correct option is (B) 2

Ordinate of point B(3, 7) = 7

Ordinate of point A(2, 5) = 5

\(\because\) 7 - 5 = 2

Hence, the ordinate of B is 2 more than the ordinate of A.

Correct option is  B) 2

Correct option is (C) (0, 0)

Both axes intersect each other at origin.

\(\therefore\) Origin lies on both axes.

Hence the point which lies on both axes is origin (0, 0).

Correct option is  C) (0, 0)

Correct option is (D) OY’

x-coordinate of point (0, -3) is O.

\(\therefore\) (0, -3) lies on negative y-axis.

i.e., (0, -3) lies on OY'.

Correct option is  D) OY’

Correct option is (C) –6

The ordinate of the point (4, -6) = y-coordinate of point (4, -6)

= - 6

Correct option is   C) – 6

Correct option is (D) (1,-1)

(-4, 2); -4 < 2 \(\Rightarrow\) abscissa < ordinate

(2, 4); 2 < 4 \(\Rightarrow\) abscissa < ordinate

(-1, 1); -1 < 1 \(\Rightarrow\) abscissa < ordinate

(1, -1); 1 > -1 and 1 - (-1) = 2

i.e., abscissa is more than its ordinate by 2.

Correct option is   D)(1,-1)

Correct option is (D) (3,-1)

Abscissa = x-coordinate of point = 3

Ordinate = y-coordinate of point = abscissa - 4

= 3 - 4 = -1

\(\therefore\) Point is (x, y) = (3, -1).

Correct option is  D) (3,-1)

(B) The answer is y = 0

(A) The answer is (a, a)

Equation of Y-axis is x = 0. 

Equation of the line parallel to the Y-axis is x = – 4. … [Given] 

∴ Distance between the Y-axis and the line x = – 4 is 0 – (- 4) … [0 > -4] 

= 0 + 4 = 4 units 

∴ The distance between the Y-axis and the line x = – 4 is 4 units. 

[Note: The question is modified as X-axis cannot be parallel to the line x = – 4.]

To find: Transformed equation of given equation when the origin (0, 0) is shifted at point (ab/(a – b), 0). 

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation. 

Since, origin has been shifted from (0, 0) to (1, 1); therefore any arbitrary point (x, y) will also be converted as (x + 1, y + 1) or (x + 1, y + 1).

(i) x² + xy – 3x – y + 2 = 0 

Substituting the value of x by x + 1 and y by y + 1, we have 

= (x + 1)² + (x + 1)(y + 1) – 3(x + 1) – (y + 1) + 2 

= 0 = x² + 1 + 2x + xy + x + y + 1 – 3x – 3 - y - 1 + 2 = 0 

= x² + xy = 0 

Hence, the transformed equation is x² + xy = 0.

(ii) x² – y² – 2x + 2y = 0 

Substituting the value of x and y by x + 1 and y + 1 respectively, we have 

= (x + 1)² – (y + 1)² – 2(x + 1) + 2(y + 1) = 0 

= x² + 1 + 2x - y² – 1 – 2y – 2x – 2 + 2y + 2 = 0 

= x² - y² = 0 

Hence, the transformed equation is x² - y² = 0.

(iii) xy – x – y + 1 = 0

Substituting the value of x and y by x + 1 and y + 1 respectively, we have 

= (x + 1)(y + 1) – (x + 1) - (y + 1) + 1 = 0 

= xy + x + y + 1 – x – 1 – y – 1 + 1 = 0 

= xy = 0 

Hence, the transformed equation is xy = 0.

(iv) xy – y² – x + y = 0 

Substituting the value of x and y by x + 1 and y + 1 respectively, we have 

= (x + 1)(y + 1) – (y + 1)² - (x + 1) + (y + 1) = 0 

= xy + x + y + 1 – y² – 1 – 2y - x – 1 + y + 1 = 0 

= xy - y² = 0 

Hence, the transformed equation is xy - y² = 0.

Given: (2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

To find: the value of p

p = (-5+11)/2 

= 6/2 

= 3

Given:

The equation x² + xy – 3x + 2 = 0

We know that the origin has been shifted from (0, 0) to (p, q)

So any arbitrary point (x, y) will also be converted as (x + p, y + q).

The new equation is:

(x + p)² + (x + p)(y + q) – 3(x + p) + 2 = 0

Upon simplification,

x² + p² + 2px + xy + py + qx + pq – 3x – 3p + 2 = 0

x² + xy + x(2p + q – 3) + y(q – 1) + p² + pq – 3p – q + 2 = 0

For no first degree term, we have 2p + q – 3 = 0 and p – 1 = 0, and

For no constant term we have p² + pq – 3p – q + 2 = 0.

By solving these simultaneous equations we have p = 1 and q = 1 from first equation.

The values p = 1 and q = 1 satisfies p² + pq – 3p – q + 2 = 0.

Hence, the point to which origin must be shifted is (p, q) = (1, 1).

(i) x² + xy – 3x – y + 2 = 0

Firstly let us substitute the value of x by x + 1 and y by y + 1

Then,

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

x² + 1 + 2x + xy + x + y + 1 – 3x – 3 – y – 1 + 2 = 0

Upon simplification we get,

x² + xy = 0

∴ The transformed equation is x² + xy = 0.

(ii) x² – y² – 2x + 2y = 0

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1)² – (y + 1)² – 2(x + 1) + 2(y + 1) = 0

x² + 1 + 2x – y² – 1 – 2y – 2x – 2 + 2y + 2 = 0

Upon simplification we get,

x² – y² = 0

∴ The transformed equation is x² – y² = 0.

(iii) xy – x – y + 1 = 0

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1) (y + 1) – (x + 1) – (y + 1) + 1 = 0

xy + x + y + 1 – x – 1 – y – 1 + 1 = 0

Upon simplification we get,

xy = 0

∴ The transformed equation is xy = 0.

(iv) xy – y² – x + y = 0

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1) (y + 1) – (y + 1)² – (x + 1) + (y + 1) = 0

xy + x + y + 1 – y² – 1 – 2y – x – 1 + y + 1 = 0

Upon simplification we get,

xy – y² = 0

∴ The transformed equation is xy – y² = 0.

Given, equation x² + xy – 3x + 2 = 0 

Let’s assume that the origin is shifted at point (p, q). 

To find: The shifted point (p, q) satisfying the question’s conditions. 

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation. 

Since, origin has been shifted from (0, 0) to (p, q); therefore any arbitrary point (x, y) will also be converted as (x + p, y + q). 

The New equation hence becomes: 

= (x + p)² + (x + p)(y + q) – 3(x + p) + 2 = 0 

= x² + p² + 2px + xy + py + qx + pq – 3x – 3p + 2 = 0 

= x² + xy + x(2p + q – 3) + y(q – 1) + p² + pq – 3p – q + 2 = 0 

For no first degree term, we have 2p + q - 3 = 0 and p – 1 = 0, 

and for no constant term we have p² + pq – 3p - q + 2 = 0. 

Solving these simultaneous equations we have p = 1 and q = 1 from first equation. And, p = 1 and q = 1 satisfies p² + pq – 3p - q + 2 = 0. 

Hence, the point to which origin must be shifted is (p, q) = (1, 1).

(b) 3x – y = 0 

Given lines are 3x – y – 3 = 0 and 3x – y + 5 = 0. 

Line parallel to the given lines can be written as 

3x – y + c = 0                              ...(i) 

Let us taken a point, say, (0, c) on (i) 

(Putting x = 0 in (i), we get y = c)

∴ \(\frac{\text{Distance of (0,c) from}\,3x-y-3}{\text{Distance of (0,c) from}\,3x-y+5}\) = \(\frac{3}{5}\)

⇒ \(\frac{\frac{|3\times0-c-3|}{\sqrt{3^2+1^2}}}{\frac{|3\times0-c+5|}{\sqrt{3^2+1^2}}}\) = \(\frac{3}{5}\)⇒ \(\frac{c+3}{-c+5}\) = \(\frac{3}{5}\)

⇒ 5c + 15 = – 3c + 15 ⇒ 8c = 0 ⇒ c = 0. 

Substituting c = 0 in (i), the required equation is 3x – y = 0.