The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th

The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th
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(b) (2, 2)Let the point be P whose abscissa = ordinate = a. ∴ P ≡ (a, a) Given, PA = PB ⇒ (a + 1)2 + a2 = a2 + (a – 5)2 ⇒ 2a2 + 2a + 1 = 2a2 – 10a + 25 ⇒ 12a = 24 ⇒ a = 2. ∴ The point is (2, 2).
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Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

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Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

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Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

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Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

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Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

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Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

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If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

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Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

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P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ