Tag: division of a line segment

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Multiple choice maths constructions mid-point formula midpoints division of a line segment

The locus of the mid point of the portion intercepted between the axes by the line $x{\,}cos\alpha+y{\,}sin{\,} \alpha=p$, where $p\inR$, is

  1. $x^2+y^2=\dfrac{4}{p^2}$
  2. $\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{4}{p^2}$
  3. $\dfrac{1}{x^2}-\dfrac{1}{y^2}=\dfrac{4}{p^2}$
  4. $\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{2}{p^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line x cos(alpha) + y sin(alpha) = p intercepts the axes at (p / cos(alpha), 0) and (0, p / sin(alpha)). The midpoint (h, k) of these intercepts is h = p / (2 cos(alpha)) and k = p / (2 sin(alpha)). Squaring and adding 1/h^2 + 1/k^2 yields 4/p^2.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

Locus of the midpoints of the intercepts between the co-ordinate Axes by the lines passing through (a, 0) does not intersect

  1. X axis

  2. Y axis

  3. Y=x

  4. Y=a
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Any line passing through (a, 0) has an intercept form x/p + y/q = 1 where it passes through (a, 0), meaning a/p + 0/q = 1, so p = a. The intercepts are (a, 0) and (0, q), so their midpoint has coordinates (h, k) = (a/2, q/2). This gives q = 2k, and since the y-intercept is q, the line intersects the y-axis at (0, 2k). However, the line cannot pass through the origin if it has distinct x and y intercepts, meaning q cannot be zero, which translates to the locus not intersecting the y-axis because h = a/2 is a vertical line parallel to the y-axis.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

I every points on the line $(a _{1}-a _{2})x+(b _{1}-b _{2}),y=c$ is equidistance from the points $(a _{1},b _{1})$  and $(a _{2},b _{2})$ then $2c=$  

  1. $a _{1}^{2}-b _{1}^{2}+a _{2}^{2}-b _{2}^{2}$
  2. $a _{1}^{2}+b _{1}^{2}+a _{2}^{2}+b _{2}^{2}$
  3. $a _{1}^{2}+b _{1}^{2}-a _{2}^{2}-b _{2}^{2}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Any point on the given line is equidistant from (a1, b1) and (a2, b2), meaning it lies on the perpendicular bisector of the segment joining those two points. The equation of the perpendicular bisector can be found by equating the squared distances from a point (x, y) to both fixed points, which yields 2(a2 - a1)x + 2(b2 - b1)y = a2^2 + b2^2 - a1^2 - b1^2. Comparing this with the given line equation (a1 - a2)x + (b1 - b2)y = c, we can equate coefficients to find that 2c equals a1^2 + b1^2 - a2^2 - b2^2.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The line equally inclined to the coordinates axes and equidistant from points A(1, -2) and B(3, 4) is

  1. x+y=2, x+y=3

  2. x-y=3, x-y=1

  3. x-y=1, x+y=3

  4. x+y=2, x-y=3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A line equally inclined to the axes has a slope of either 1 or -1, meaning its equation is of the form x - y = c or x + y = c. Since it is equidistant from points A(1, -2) and B(3, 4), it must pass through the midpoint of AB or be parallel to the line segment AB. Testing the options reveals that the pair x - y = 1 and x + y = 3 satisfies the distance conditions from the midpoint and the slope requirements.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The midpoint of the interval in which $x^{2}-2(\sqrt{-x})^{2}-3<0$ is satisfied, is

  1. $\dfrac{-3}{2}$
  2. $-2$
  3. $\dfrac{1}{2}$
  4. $\dfrac{-3}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression sqrt(-x)^2 is defined for x <= 0 and equals -x. The inequality is x^2 - 2(-x) - 3 < 0 => x^2 + 2x - 3 < 0 => (x+3)(x-1) < 0. Since x <= 0, the interval is [-3, 0]. The midpoint is (-3+0)/2 = -3/2.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

A tangent to the circle $x^{2}+y^{2}=a^{2}$ meets the axes at points A and B. The locus of the mid point of AB is 

  1. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{a^{2}}$
  2. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{a^{2}}$
  3. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=4a^{2}$
  4. $\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{a^{2}}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the tangent to the circle x^2 + y^2 = a^2 be written in intercept form as x/p + y/q = 1. Since it is tangent to the circle centered at the origin with radius a, the perpendicular distance from the origin to the line equals a, which gives 1/p^2 + 1/q^2 = 1/a^2. The intercepts on the axes are A(p, 0) and B(0, q), so the midpoint (h, k) of AB is (p/2, q/2), meaning p = 2h and q = 2k. Substituting these into the tangent condition yields 1/(4h^2) + 1/(4k^2) = 1/a^2, which simplifies to 1/x^2 + 1/y^2 = 4/a^2.