Advertisements
Advertisements
Question
Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).
Advertisements
Solution
We know that the y and z coordinates of the point on the x-axis are 0.
So, let the required point be C (x, y, z)
Now, CA = CB
\[\sqrt{\left( 3 - x \right)^2 + \left( 2 - 0 \right)^2 + \left( 2 - 0 \right)^2} = \sqrt{\left( 5 - x \right)^2 + \left( 5 - 0 \right)^2 + \left( 4 - 0 \right)^2}\]
\[ \Rightarrow 9 - 6x + x^2 + 4 + 4 = 25 - 10x + x^2 + 25 + 16\]
\[ \Rightarrow 17 - 6x + x^2 = 66 - 10x + x^2 \]
\[ \Rightarrow 4x = 49\]
\[ \Rightarrow x = \frac{49}{4}\]
Hence, the required point is\[\left( \frac{49}{4}, 0, 0 \right)\]
APPEARS IN
RELATED QUESTIONS
Coordinate planes divide the space into ______ octants.
Name the octants in which the following points lie:
(4, –3, 5)
Name the octants in which the following points lie:
(7, 4, –3)
Name the octants in which the following points lie:
(2, –5, –7)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(–2, 3, 4) in the yz-plane.
Find the image of:
(–5, 4, –3) in the xz-plane.
Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.
Find the distances of the point P(–4, 3, 5) from the coordinate axes.
Find the coordinates of the point which is equidistant from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
Verify the following:
(0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.
Find the point on y-axis which is at a distance of \[\sqrt{10}\] units from the point (1, 2, 3).
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are
The perpendicular distance of the point P (6, 7, 8) from xy - plane is
The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is
If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.
Find the direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).
The x-coordinate of a point on the line joining the points Q(2, 2, 1) and R(5, 1, –2) is 4. Find its z-coordinate.
Find the image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.
Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0
Show that the points `(hati - hatj + 3hatk)` and `3(hati + hatj + hatk)` are equidistant from the plane `vecr * (5hati + 2hatj - 7hatk) + 9` = 0 and lies on opposite side of it.
Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = (5hati - 4hatj + 6hatk) + lambda(3hati + 7hatj - 2hatk)`.
